FUNDAMENTAL PARTICLES
Atoms are made upessentially, of three fundamental particles, which differ in mass and electric charge as follows:
Electron  Proton  Neutron  
Symbol  e or e–  p  n 
Approximate relative mass  1/1836  1  1 
Approximate relative charge  −1  +1  0 
Mass in kg  9.109534 × 10−31  1.6726485 × 10−27  1.6749543 × 10−27 
Mass in amu  5.4858026 × 10−4  1.007276471  1.008665012 
Actual charge/C  1.6021892 × 10−19  1.6021892 × 10−19  0 
The atomic mass unit (amu) is 1/12th of the mass of an individual atom of 6C12, i.e.
1.660565 × 10−27 kg. The neutron and proton have approximately equal masses of 1 amu and the electron is about 1836 times lighter; its mass can sometimes be neglected as an approximation. The electron and proton have equal, but opposite, electric charges; the neutron is not charged.
The existence of electrons in atoms was first suggested, by J.J. Thomson, as a result of experimental work on the conduction of electricity through gases at low pressures, which produces cathode rays and xrays, and a study of radioactivity by Becquerel, the Curies and Rutherford.
An atom is electrically neutral, and if it contains negatively charged electrons it must also contain some positively charged particles, and the supposition that they existed within atoms came about as a result of Rutherford’s experiments in which he bombarded elements with the – rays and
rays were given off by radioactive elements. The neutron was discovered in 1932 by James Chadwick by bombarding beryllium with rays.
DISCOVERY OF ELECTRON: CATHODE RAYS
During the latter half of the nineteenth century, it was found that while normally dry gases do not conduct an electric current, they do so under very low pressure and then patches of light are seen. The passage of electricity through gases as studied by a number of physicists, particularly by Faraday, Davy, Crookes and J.J. Thomson.
When a current of high voltage (10,000 volts) is passed through a gas of air kept at a very low pressure (0.01 – 0.03 mm) blue rays are seen emerging from the case. These rays are called “Cathode Rays”.
Some of the important properties of the cathode rays studied by Sir J.J. Thomson and others are given below:
 Cathode rays come out at right angles to the surface of the cathode and move in
straight lines.  Their path is independent on the position of the anode.
 They produce phosphorescence on certain salts like ZnS and fluorescence on glass.
 They blacken photographic plates.
 The rays pass through thin sheet of metals. If the metal sheet is too thick to be penetrated the rays cast a shadow.
 They produce X−ray when they strike a metal.
 The rays ionize a gas through which they pass.
 They heat a substance on which they fall.
 They rotate a light wheel placed in their paths. This shows that cathode rays contain material particles having both mass and velocity.
 The mass of a particle present in cathode rays is found to be 1/1837 of H−atom. This shows that the particle is of sub−atomic nature.
 Cathode rays are deflected by a magnetic or an electric field showing the particle to be electrically charged, the direction of deflection shows that they are negatively charged.
 Cathode rays contain the smallest unit of negative charge.
 No cathode ray was produced when the tube was completely evacuated.
 Different gases produce same cathode rays as they have the same e/m (charge/mass) ratio. This indicates that the particles present in cathode rays are fundamental constituent of all matter.
Sir J.J. Thomson named these negatively charged sub−atomic particles as electron.
“A sub−atomic particle which is a fundamental constituent of all matter having a mass 1/1837th of a H−atom and which carries the smallest unit of negative charge is called an electron”.
Determination of Velocity and Charge/mass (e/m) ratio of Electrons:
Sir J.J. Thomson (1897) extended the cathode ray experiment for the determination of velocity of electrons and their charge/mass ratio, The value of e/m for an electron = 1.76 × 108 C/g.
For the H+ ion (proton), e/m = 96500/1.008 C/g.
Millikan’s Oil Drop Method: Determination of Charge on an Electron:
In 1909, Millikan measured the charge on an electron by his oil drop method. In this method a spray of oil droplets is produced by an atomizer, some of which pass through an opening into a viewing chamber, where we can observe them with a microscope. Often these droplets have an electric charge, which is picked up from the friction forming the oil droplets. A droplet may have one or more additional electrons in it, giving it a negative charge.
As the droplet falls to the bottom of the chamber, it passes between two electrically charged plates. The droplet can be suspended between them; we adjust the voltage in the plates so that the electrical attraction upward just balances the force of gravity downward. We then use the voltage needed to establish this balance to calculate the mass – to charge ratio for the droplet. Because we already know the mass of the droplet we can find the charge on it.
Millikan’s found that the charge on all droplets could be expressed as whole number multiples of e, where the value of e is 1.602×1019 C. By combining e/m. ratio and ‘e’ we calculate mass of the electron
= 9.104×1031 kg
This very small value shows that the electron is a subatomic particle. Thus charge on an electron = 1.602 × 10–19C.
DISCOVERY OF PROTON: POSITIVE RAYS OR CANAL RAYS
Atoms are electrically neutral. Hence after the discovery of the negatively charged constituent (electron) of an atom, attempts were made to discover the positively charged counterpart of electrons. By using a discharge tube containing a perforated cathode. Goldstein (1886) found that some rays passed through these holes in a direction opposite to that of the cathode rays.
These are called the positive rays or canal rays. J.J. Thomson (1910) measured their charge by mass ratio from which he was able to deduce that these contain positive ions. Their
properties are:
 They are positively charged.
 The positive charge is either equal to or whole number multiple of the charge on an electron.
 When hydrogen gas was filled in the discharge tube the positive charge on the positive rays was equal to the negative charge on an electron, and the mass was less than the hydrogen atom.
 Unlike cathode rays the properties of positive rays are characteristics of the gas in
the tube.  The deflection of positive rays under the influence of an electric or magnetic field is smaller than that of the cathode rays for the same strength of field. This shows that the positive rays have a greater mass than that of electrons.
 The mass of the positive rays depends on the atomic weights or molecular weights of the gases in the discharge tube. The charge/mass ratio also varies because the change in positive charge on the rays. It may be either equal to or integral multiple of the charge on an electron.
 The lightest of all particles identified in positive rays from different elements was one with a mass very slightly less than that of hydrogen atom (or nearly equal to H−atom). The lightest positively charged particle is called a proton (P or P+). Positive rays are atomic or molecular resides from which some electrons have been removed. The removed electrons constitute the cathode rays and the positive residues form the positive or canal rays.
Positive Rays  Cathode Rays  
H  H+  e– 
O ⎯→  O+  e– 
O2 ⎯→  O2+  e– 
O2 ⎯→  O22+  2e– 
The mass of a proton is very slightly less than that of a H−atom. This shows that protons are sub−atomic particles. Protons are fundamental constituent of matter because positive rays are produced by all substances.
“A sub−atomic particle, which is a fundamental constituent of all matter having a mass slightly less than that of H−atom and which carries a positive charge equal in magnitude to the charge on an electron, is called a proton”. A proton is denoted by p or p+ of +1p.
Comparison of Positive (Canal) Rays and Cathode Rays:
Properties  Cathode Rays  Canal Rays 
Sign of Charge  Negative  Positive 
Mag. of Charge  Always –1  Mostly +1, but also +2, +3… 
Mass  Definite value  Variable, depends on ions 
e/m  Definite value  Variable, depends on ions 
DISCOVERY OF NEUTRON
After the discovery of electrons and protons. Rutherford (1920) had predicted the existence of a neutral fundamental particle. In 1932, Chadwick bombarded the element Beryllium with α−particles and noticed the emission of a radiation having the following characteristics.
 The radiation was highly penetrating.
 The radiation was unaffected by magnetic and electric fields which show that it is electrically neutral.
 It was found to have approximately the same mass as the protons.
The name ‘neutron’ was given to this sub−atomic particle. It is denoted by n or. Bombardment of beryllium by α−particles results in the formation of carbon and neutrons are emitted.
Be
At present there are a number of evidences which confirm that like electron, proton and neutron is also a fundamental constituent of atoms (a single exception is atom which does not contain any neutron)
Mass of a neutron is 1.008930 amu (1.6753 × 10–24g or 1.6753 × 10–27 kg)
Neutron “A sub−atomic particle, which is a fundamental constituent of matter having mass approximately equal to the hydrogen atom and which is electrically neutral, is called a neutron”.
Illustration 1. The neutron is attracted towards
(A) Positive charged particles
(B) Negative charged particles
(C) Not attracted by any charge
(D) None of these
Solution: Neutron is an uncharged particle. Hence (C) is correct.
ATOMIC TERMS
Nuclide: Various species of atoms in general.
Nucleons: Subatomic particles in the nucleus of an atom, i.e., protons and neutrons.
Isotopes: Atoms of an element with the same atomic number but different mass number.
Mass number (A): Sum of the number of protons and neutrons, i.e., the total number of nucleons,
Atomic number (Z): The number of protons in the nucleus of an atom. This, when subtracted from A, gives the number of neutrons.
Isobars: Atoms, having the same mass numbers but different atomic numbers, e.g.. 15P32 and 16S32.
Isotones: Atoms having the same number of neutrons but different number of protons or mass number, e.g.,
Isoelectronic species: Atoms molecules or ions having the same number of electrons, e.g., N2, CO, CN–.
Nuclear isomers: Atoms with the same atomic and mass numbers but different radioactive properties, e.g., uranium X (half life 1.4 min) and uranium Z (half life 6.7 hours).
Atomic mass unit: Exactly equal to 1/12th of the mass of 6C12 atom.
(a.m.u.): 1 a.m.u. = 1.66 × 10–24 g ≈ 931.5 MeV
Illustration 2. The ion that is isoelectronic with CO is
(A) CN– (B) O2+
(C) O2– (D) N2+
Solution: Both CO and CN− have 14 electrons. Hence (A) is correct
Illustration 3. Find the number of neutrons in a neutral atom having atomic mass 23 and number of electrons 11.
Solution: Number of protons = number of electrons = 11
Number of neutrons = atomic mass – number of protons = 23 – 11 = 12
Illustration 4. How many protons, electrons and neutrons are present in 0.18 g?
Solution: No. of neutrons in one atom = (30 – 15) = 15
0.18 g = = 0.006 mole
Now, number of atoms in 0.006 mole = 0.006 × 6.02 × 1023
Number of electrons in 0.006 mole of =Number of protons in 0.006 mole = 15 × 0.006 × 6.02 × 1023 = 5.418 × 1022 and number of neutrons = 5.418 × 1022
Exercise 1.
(i) Find the mass number of a uninegative ion having 10 neutrons and
10 electrons in one mole of the ions.
(ii) Find the total number and total mass of protons present in 34 mg of NH3.
ATOMIC MODELS
We know the fundamental particles of the atom. Now let us see, how these particles are arranged in an atom to suggest a model of the atom.
Thomson’s Model:
J.J. Thomson, in 1904, proposed that there was an equal and opposite positive charge enveloping the electrons in a matrix. This model is called the plum – pudding model after a type of Victorian dessert in which bits of plums were surrounded by matrix of pudding.
This model could not satisfactorily explain the results of scattering experiment carried out by Rutherford who worked with Thomson.
Rutherford’s Model:
α– particles emitted by radioactive substance were shown to be dipositive Helium ions (He++) having a mass of 4 units and 2 units of positive charge.
Rutherford allowed a narrow beam of α–particles to fall on a very thin gold foil of thickness of the order of 0.0004 cm and determined the subsequent path of these particles with the help of a zinc sulphide fluorescent screen. The zinc sulphide screen gives off a visible flash of light when struck by an α particle, as ZnS has the remarkable property of converting kinetic energy of α particle into visible light. [For this experiment, Rutherford specifically used α particles because they are relatively heavy resulting in high momentum].
Observation:
 Majority of the α–particles pass straight through the gold strip with little or no deflection.
 Some α–particles are deflected from their path and diverge.
 Very few α–particles are deflected backwards through angles greater than 90°.
 Some were even scattered in the opposite direction at an angle of 180° [Rutherford was very much surprised by it and remarked that “It was as incredible as if you fired a 15–inch shell at a piece of tissue paper and it came back and hit you”]. There is far less difference between air and bullet than there is between gold atoms and αparticle assuming of course that density of a gold atom is evenly distributed. The distance of nucleus from where the α – particle returns back through 180° is called distance of closet approach and is given by
Conclusions:
 The fact that most of the α – particles passed straight through the metal foil indicates the most part of the atom is empty.
 The fact that few α – particles are deflected at large angles indicates the presence of a heavy positively charge body i.e., for such large deflections to occur α – particles must have come closer to or collided with a massive positively charged body.
 The fact that one in 20,000 have deflected at 180° backwards indicates that volume occupied by this heavy positively charged body is very small in comparison to total volume of the atom.
Illustration 5. An α – particle is traveling towards gold nuclei returns back through 10−10 m from it. What is the velocity of the α – particle. [Given 1 amu = 1.66 × 10−27 kg, atomic mass of He = 4 and gold = 79 and = 9 × 109 Nm2C−2]
Solution: We known that
Now one He atom has charge (q1) = 2e
One gold atom has charge (q2) = 79e
Putting these values we get
∴ v = 3.311 × 105 m/s
Conclusions of αScattering Experiment:
On the basis of the above observation, and having realized that the rebounding αparticles had met something even more massive than themselves inside the gold atom, Rutherford proposed an atomic model as follows.
 All the +ve charge and nearly the total mass of an atom is present in a very small region at the centre of the atom. The atom’s central core is called nucleus.
 The size of the nucleus is very small in comparison to the size of the atom. Diameter of the nucleus is about 10–13cm while the atom has a diameter of the order of 10–8 cm. So, the size of atom is 105 times more than that of nucleus.
 Most of the space outside the nucleus is empty.
 The electrons, equal in number to the net nuclear positive charge, revolve around the nucleus with fast speed just like planets around the sun.
 The centrifugal force arising due to the fast speed of an electron balances the coulombic force of attraction of the nucleus and the electron remains stable in its path. Thus according to him atom consists of two parts (a) nucleus and (b) extra nuclear part.
Defects in Rutherford’s Atomic Model:
 Position of electrons: The exact positions of the electrons from the nucleus are not mentioned.

It was calculated that the electron should fall into the nucleus in less than 10–8 sec. But it is known that electrons keep moving outside the nucleus.
To solve this problem Neils Bohr proposed an improved form of Rutherford’s atomic model.
Before going into the details of Neils Bohr model we would like to introduce you some important atomic terms.
Illustration 6. Prove that density of the nucleus is constant.
Solution: Radius of the nucleus = 1.33 × 10–13 × A1/3 cm, where A is the mass number
= 1.33 × 10–11 × A1/3 m
Density of nucleus =
== kg/m3 = constant
Thus density of nucleus is constant, independent of the element under consideration.
SOME IMPORTANT CHARACTERISTICS OF A WAVE
A wave is a sort of disturbance which originates from some vibrating source and travels outward as a continuous sequence of alternating crests and troughs. Every wave has five important characteristics, namely, wavelength (λ), frequency (ν), velocity (c), wave number and amplitude (a).
Electronic Magnetic Radiation:
Ordinary light rays, X–rays,γ–rays, etc. are called electromagnetic radiations because similar waves can be produced by moving a charged body in a magnetic field or a magnet in an electric field. These radiations have wave characteristics and do not require any medium for their propagation.
 Wavelength (λ): The distance between two neighbouring troughs or crests is known as wavelength. It is denoted by λ and is expressed in cm, m, nanometers (1 nm =10–9 m) or Angstrom (1 Å=10–10 m).
 Frequency (ν): The frequency of a wave is the number of times a wave passes through a given point in a medium in one second. It is denoted by ν(nu) and is expressed in cycles per second (cps) or hertz (Hz) 1Hz = 1cps.
The frequency of a wave is inversely proportional to its wave length (λ)
ν ∝ or ν =
 Velocity: The distance travelled by the wave in one second is called its velocity. It is denoted by c and is expressed in cm sec–1.
c = νλ or λ =
 Wave number: It is defined as number of wavelengths per cm. It is denoted by and is expressed in cm–1.
= or =
 Amplitude: It is the height of the crest or depth of the trough of a wave and is denoted by a. It determines the intensity or brightness of the beam of light.
 Electromagnetic Spectrum: The arrangement of the various types of electromagnetic radiation in order of increasing or decreasing wavelengths or frequencies is known as electromagnetic spectrum.
Wavelengths of Electromagnetic Radiations:
Electromagnetic radiations  Wave length (Å) 
Radio waves  3×1014 to 3 ×107 
Micro waves  3×109 to 3 ×106 
Infrared (IR)  6×106 to 7600 
Visible 
7600 to 3800 
Ultra violet (UV)  3800 to 150 
X–rays  150 to 0.1 
Gamma rays  0.1 to 0.01 
Cosmic rays  0.01 to zero 
Illustration 7. Find out the longest wavelength of absorption line for hydrogen gas containing atoms in ground state.
Solution:
For longest wavelength ΔE should be smallest, i.e. transition occurs from
n = 1 to n = 2
i.e. = 109673 cm–1 × 12
∴ = 109673 × cm–1
∴ λ = = 1.2157 × 10–5 cm = 121.6 nm
Illustration 8. Calculate the energy in kJ per mole of electronic charge accelerated by a potential of 1 volt.
Solution: Energy in joules = charge in coulombs × potential difference in volt
= 1.6 ×10–19 × 6.02 × 1023 × 1
= 9.632 × 104 J or 96.32 kJ
Exercise 2.
(i) What is the ratio between energies of two radiations one with a wavelength of 6000 Å and other with 2000 Å.
(ii) Find the frequency and wave number of a radiation having wavelength of 1000 Å.
ATOMIC SPECTRUM
If the atom gains energy the electron passes from a lower energy level to a higher energy level, energy is absorbed that means a specific wave length is absorbed. Consequently, a dark line will appear in the spectrum. This dark line constitutes the absorption spectrum.
If the atom loses energy, the electron passes from higher to a lower energy level, energy is released and a spectral line of specific wavelength is emitted. This line constitutes the emission spectrum.
Types of Emission Spectra:
 Continuous spectra: When white light from any source such as sun or bulb is analyzed by passing through a prism, it splits up into seven different wide bands of colour from violet to red (like rainbow). These colour are so continuous that each of them merges into the next. Hence the spectrum is called as continuous spectrum.
 Line spectra: When an electric discharge is passed through a gas at low pressure light is emitted. If this light is resolved by a spectroscope, It is found that some isolated coloured lines are obtained on a photographic plate separated from each other by dark spaces. This spectrum is called line spectrum. Each line in the spectrum corresponds to a particular wavelength. Each element gives its own characteristic spectrum.
PLANCK’S QUANTUM THEORY
When a black body is heated, it emits thermal radiations of different wavelengths or frequency. To explain these radiations, Max Planck put forward a theory known as Planck’s quantum theory. The main points of quantum theory are
 Substances radiate or absorb energy discontinuously in the form of small packets or bundles of energy.
 The smallest packet of energy is called quantum. In case of light the quantum is known as photon.
 The energy of a quantum is directly proportional to the frequency of the radiation. E ∝ ν (or) E = hν where ν is the frequency of radiation and h is Planck’s constant having the value 6.626 × 10–27 erg – sec or 6.626 × 10–34 J–sec.
 A body can radiate or absorb energy in whole number multiples of a
quantum hν, 2hν,3hν………..nhν. where ‘n’ is the positive integer.  Neils Bohr used this theory to explain the structure of atom.
Illustration 9. The wave number of a radiation is 400 cm−1. Find out its
(a) Wavelength (b) Frequency
(c) J per photon (d) kcal per mol of photons
(e) kJ per mol of photons
Solution: (a) or λ = = = 2.5 × 10–3 cm
(b) ν = = = 3 × 10–10 cm/s × 400 cm–1 =
(c) Ephoton = hν = =
= 6.626 × 10–34 Js × 3 × 108 cm/s × 400 cm–1
= 7.95 × 10–21J
(d) Ephoton = 7.95 × 10–21J
For 1 mol of photon energy = 7.95 × 10–21J × 6.022 × 1023 mol–1
= (4.7875 × 103 J mol–1) (1 kcal/4184 J)
= 1.14 kcal mol–1
(e) E = (4.7875 × 103 J mol–1) (1 kJ/1000J)
= 4.7875 kJ mol–1
Illustration 10. A near ultraviolet photon of 300 nm is absorbed by a gas and then re−emitted as two photons. One photon is red with wavelength 760 nm. What would be the wavelength of the second photon?
Solution: E =
Wavelength of the 1st photon = λ1
∴ E1 =
And that of second photon = λ2
∴ E2 =
ETotal = E1 + E2= emitted energy
λ2 = wavelength of the second photon = 496 nm
BOHR’S ATOMIC MODEL
Bohr developed a model for hydrogen atom and hydrogen like one–electron species (hydrogenic species). He applied quantum theory in considering the energy of an electron bound to the nucleus.
Important Postulates:
 An atom consists of a dense nucleus situated at the centre with the electron revolving around it in circular orbits without emitting any energy. The force of attraction between the nucleus and an electron is equal to the centrifugal force of the moving electron.
 Of the finite number of circular orbits around the nucleus, an electron can revolve only in those orbits whose angular momentum (mvr) is an integral multiple of factor
mvr =
where, m = mass of the electron
v = velocity of the electron
n = orbit number in which electron is present
r = radius of the orbit
 As long as an electron is revolving in an orbit it neither loses nor gains energy. Hence these orbits are called stationary states. Each stationary state is associated with a definite amount of energy and it is also known as energy levels. The greater the distance of the energy level from the nucleus, the more is the energy associated with it. The different energy levels are numbered as 1,2,3,4, ( from nucleus onwards) or K,L,M,N etc.
 Ordinarily an electron continues to move in a particular stationary state without losing energy. Such a stable state of the atom is called as ground state or normal state.
 If energy is supplied to an electron, it may jump (excite) instantaneously from lower energy (say 1) to higher energy level (say 2, 3, 4, etc.) by absorbing one or more quanta of energy. This new state of electron is called as excited state. The quantum of energy absorbed is equal to the difference in energies of the two concerned levels.
Since the excited state is less stable, atom will lose it’s energy and come back to the ground state.
Energy absorbed or released in an electron jump, (ΔE) is given by
ΔE = E2 – E1 = hν
where E1 and E2 are the energies of the electron in the first and second energy levels, and ν is the frequency of radiation absorbed or emitted.
Note: If the energy supplied to hydrogen atom is less than 13.6 eV, it will accept or absorb only those quanta which can take it to a certain higher energy level i.e., all those photons having energy less than or more than a particular energy level will not be absorbed by hydrogen atom. But if energy supplied to hydrogen atom is more than 13.6 eV then all photons are absorbed and excess energy appears as kinetic energy of emitted photo electron.
Radius and Energy Levels of Hydrogen Atom:
Consider an electron of mass ‘m’ and charge ‘e’ revolving around a nucleus of charge Ze (where, Z = atomic number and e is the charge of the proton) with a tangential velocity v. r is the radius of the orbit in which electron is revolving.
By Coulomb’s Law, the electrostatic force of attraction between the moving electron
and nucleus is
Coulombic force =
K = (where ∈o is permittivity of free space)
K = 9 ×109 Nm2 C–2
In C.G.S. units, value of K = 1 dyne cm2 (esu)–2
The centrifugal force acting on the electron is
Since the electrostatic force balances the centrifugal force, for the stable electron orbit.
= … (1)
(or) v2 = … (2)
According to Bohr’s postulate of angular momentum quantization, we have
mvr =
v =
v2 = … (3)
Equating (2) and (3)
Solving for r we get r =
Where n = 1, 2, 3 – – – – – ∞
Hence only certain orbits whose radii are given by the above equation are available for the electron. The greater the value of n, i.e., farther the energy level from the nucleus the greater is the radius.
The radius of the smallest orbit (n=1) for hydrogen atom (Z=1) is ro.
ro = = = 5.29 ×10–11 m = 0.529 Å
Radius of nth orbit for an atom with atomic number Z is simply written as
rn = 0.529 × Å
Calculation of Energy of an Electron:
The total energy, E of the electron is the sum of kinetic energy and potential energy.
Kinetic energy of the electron = ½ mv2
Potential energy =
Total energy = 1/2 mv2 – … (4)
From equation (1) we know that
=
∴ ½ mv2 =
Substituting this in equation (4)
Total energy (E) =– =
Substituting for r, gives us
E = where n = 1, 2, 3……….
This expression shows that only certain energies are allowed to the electron. Since this energy expression consist of so many fundamental constant, we are giving you the following simplified expressions.
E = –21.8 ×10–12 × erg per atom
= –21.8 ×10–19 × J per atom
= –13.6 × eV per atom
(1eV = 3.83 ×10–23 kcal
1eV = 1.602 ×10–12 erg
1eV = 1.602 ×10–19J)
E = –313.6 ×kcal / mole (1 cal = 4.18 J)
The energies are negative since the energy of the electron in the atom is less than the energy of a free electron (i.e., the electron is at infinite distance from the nucleus) which is taken as zero. The lowest energy level of the atom corresponds to n=1, and as the quantum number increases, E becomes less negative.
When n = ∞, E = 0, which corresponds to an ionized atom i.e., the electron and nucleus are infinitely separated.
H ⎯→ H++ e– (ionization).
Calculation of Velocity:
We know that
mvr =; v =
By substituting for r we get
v =
Where except n and Z all are constants
v = 2.18 ×108 cm/sec.
Further application of Bohr’s work was made, to other one electron species (Hydrogenic ion) such as He+ and Li2+. In each case of this kind, Bohr’s prediction of the spectrum was correct.
Illustration 11. The velocity of electron in the second orbit of will be
(A) (B)
(C) (D) None of these
Solution: . Hence (A) is correct.
Illustration 12. Calculate the velocity of an electron in Bohr’s first orbit of hydrogen atom
(Given r = 0.53 × 10–10 m)
Solution: According to Bohr’s theory
mvr =
r = 0.53 × 10–10m
n = 1, h = 6.62 × 10–34kg m2 s–1, m = 9.1 × 10–31 kg
v = = 2.18 × 106 m/s
Illustration 13. The velocity of e− in first Bohr’s orbit is 2.17 × 106 m/s. Calculate the velocity in 3rd orbit of He2+ ion.
Solution: vn =
Hence v = = 1.45 × 106 m/s
HYDROGEN ATOM
If an electric discharge is passed through hydrogen gas taken in a discharge tube under low pressure, and the emitted radiation is analyzed with the help of spectrograph, it is found to consist of a series of sharp lines in the UV, visible and IR regions. This series of lines is known as line or atomic spectrum of hydrogen. The lines in the visible region can be directly seen on the photographic film.
Each line of the spectrum corresponds to a light of definite wavelength. The entire spectrum consists of six series of lines, each series, known after their discoverer as the Balmer, Paschen, Lyman, Brackett, Pfund and Humphrey series. The wavelength of all these series can be expressed by a single formula.
= R
Where, = wave number
λ = wave length
R = Rydberg constant (109678 cm–1)
n1 and n2 have integral values as follows
Series n1 n2 Main spectral lines
Lyman 1 2, 3, 4, etc Ultra – violet
Balmer 2 3, 4, 5 etc Visible
Paschen 3 4, 5, 6 etc Infra – red
Brackett 4 5, 6, 7 etc Infra – red
Pfund 5 6, 7, etc Infra – red
Note: All lines in the visible region are of Balmer series but reverse is not true. i.e., all Balmer lines will not fall in visible region]
The pattern of lines in atomic spectrum is characteristic of hydrogen.
Illustration 14. Find the wavelength of a spectral line produced when an electron in the Hatoms jumps from 4th level to 2nd level.
Solution:
⇒
= 4863
Merits of Bohr’s Theory:
 The experimental value of radii and energies in hydrogen atom are in good agreement with that calculated on the basis of Bohr’s theory.
 Bohr’s concept of stationary state of electron explains the emission and absorption spectra of hydrogen like atoms.
 The experimental values of the spectral lines of the hydrogen spectrum are in close agreement with that calculated by Bohr’s theory.
Limitations of Bohr’s Theory
 It does not explain the spectra of atoms having more than one electron.
 Bohr’s atomic model failed to account for the effect of magnetic field (Zeeman effect) or electric field (Stark effect) on the spectra of atoms or ions. It was observed that when the source of a spectrum is placed in a strong magnetic or electric field, each spectral line further splits into a number of lines. This observation could not be explained on the basis of Bohr’s model.
 De Broglie suggested that electrons like light have dual character. It has particle and wave character. Bohr treated the electron only as particle.
 Another objection to Bohr’s theory came from Heisenberg’s Uncertainty Principle. According to this principle “It is impossible to determine simultaneously the exact position and momentum of a small moving particle like an electron”. The postulate of Bohr, that electrons revolve in well defined orbits around the nucleus with well defined velocities is thus not tenable.
Illustration 15. The radii of two of the first four Bohr orbits of the hydrogen atom are in the ratio
1 : 4. The energy difference between them may be
(A) 0.85 eV (B) 2.55 eV
(C) 3.40 eV (D) 8.20 eV
Solution: and. Hence (B) is correct.
Illustration 16. An electron in a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum required for its escape from the atom. What is the wavelength of the emitted electron? (me = 9.11 × 10–31 kg)
(A) 4.7 Å (B) 4.7 nm
(C) 9.4 Å (D) 9.40 nm
Solution: Since 13.6 eV is need for ionisation
Total energy absorbed =
∴ 6.8 eV is converted into K.E.
= Å
Hence (A) is correct.
Illustration 17. The observation that electrons can be diffracted is an evidence that electrons
(A) have particle properties (B) have wave properties
(C) are emitted by atoms (D) are absorbed by ions
Solution: Diffraction prove wave character of electron. Hence (B) is correct.
Illustration 18. A series of lines in the spectrum of atomic hydrogen lies at wavelengths 656.46, 482.7, 434.17, 410. 29 nm. What is the wavelength of next line in this series?
Solution: The given series of lines are in the visible region and thus appears to be Balmer series
Therefore n1 = 2 and n2=? for next line
If λ = 410.29 ×10–7 cm and n1 = 2
n2 may be calculated for the last line
= R
= 109673
n2 = 6
Thus next line will be obtained during the jump of electron from 7th to 2nd shell i.e,
= R = 109673
λ = 397.2 ×10–7 cm = 397.2 nm
Illustration 19. Find the energy released in (ergs) when 2.0 gm atom of hydrogen undergo transition giving spectral line of lowest energy in visible region of its atomic spectrum.
Solution: ΔE = 2.178 × 10–18 (Z2)
For visible photon, n1 = 2
For lowest energy transition n2 = 3
⇒ ΔE = 2.178 × 10–18 × 12 J/atom
For 2.0g atom
ΔE = (2 × 6.023 × 1023) 2.178 × 10–18 joules
= 3.63 × 105J = 3.63 × 1012 ergs
Illustration 20. Which hydrogen like ionic species has wavelength difference between the first line of Balmer and first line of Lyman series equal to 59.3 × 10–9 m? Neglect the reduced mass effect.
Solution: Wave number of first Balmer line of an species with atomic number Z is given by
; Similarly wave number of of first Lyman line is given by
= RZ2 =;
∴ λ′ – λ = = =
∴Z2 = = 9
or Z = 3
∴ ionic species is Li2+
Exercise 3.
(i) Calculate λ of the radiations when the electron jumps from III to II orbit of hydrogen atom. The electronic energy in II and III Bohr orbit of hydrogen atoms are –5.42 ×10–12 and –2.41 ×10–12 erg respectively.
(ii) In a hydrogen atom, the energy of an electron in first Bohr’s orbit is –13.12 × 105 J mol–1. What is the energy required for its excitation to Bohr’s
second orbit.
(iii) What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n =? What is the colour corresponding to this wavelength? (Rydberg constant = 109,677 cm−1)
QUANTUM NUMBERS
An atom contains large number of shells and subshells. These are distinguished from one another on the basis of their size, shape and orientation (direction) in space. The parameters are expressed in terms of different numbers called quantum numbers.
Quantum numbers may be defined as a set of four numbers with the help of which we can get complete information about all the electrons in an atom. It tells us the address of the electron i.e., location, energy, the type of orbital occupied and orientation of that orbital.
 Principal quantum number (n): It tells the main shell in which the electron resides, the approximate distance of the electron from the nucleus and energy of that particular electron. It also tells the maximum number of electrons that a shell can accommodate is 2n2, where n is the principal quantum number.
Shell K L M N
Principal quantum number (n) 1 2 3 4
Maximum number of electrons 2 8 18 32
 Azimuthal or angular momentum quantum number: This represents the number of subshells present in the main shell. These subsidiary orbits within a shell will be denoted as 0, 1, 2, 3, 4,… or s, p, d, f… This tells the shape of the subshells. The orbital angular momentum of the electron is given as or for a particular value of ‘n’. For a given value of n values of possible vary from 0 to n – 1.
 The magnetic quantum number (m): An electron due to its angular motion around the nucleus generates an electric field. This electric field is expected to produce a magnetic field. Under the influence of external magnetic field, the electrons of a subshell can orient themselves in certain preferred regions of space around the nucleus called orbitals. The magnetic quantum number determines the number of preferred orientations of the electron present in a subshell. The values allowed depends on the value of l, the angular momentum quantum number, m can assume all integral values between – to + including zero. Thus m can be –1, 0, +1 for = 1. Total values of m associated with a particular value of is given by 2+ 1.
 The spin quantum number (s): Just like earth which not only revolves around the sun but also spins about its own axis, an electron in an atom not only revolves around the nucleus but also spins about its own axis. Since an electron can spin either in clockwise direction or in anticlockwise direction, therefore, for any particular value of magnetic quantum number, spin quantum number can have two values, i.e., +1/2 and –1/2 or these are represented by two arrows pointing in the opposite directions, i.e., ↑ and ↓. When an electron goes to a vacant orbital, it can have a clockwise or anti clockwise spin i.e., +1/2 or –1/2. This quantum number helps to explain the magnetic properties of the substances.
Illustration 21. If the principal quantum number n has a value of 3, what are permitted values of the quantum numbers ‘’ and ‘m’?
Solution: If principal quantum number (n) has the value of 3, subsidary quantum number () will also have three values i.e. 0, 1 and 2 and magnetic quantum number (m) will have n2, i.e. 32 i.e. 9 values in all and they are designated as under:
n = 3 = 0 m = 0 (s)
= 1 m = – 1 (p)
m = 0
m = + 1
= 2 m = – 2 (d)
m = – 1
m = 0
m = +1
m = +2
Illustration 22. In which orbital the electron will reside if it has the following values of quantum numbers.
(i) n = 3,
(ii) n = 4,
Solution: (i) for p subshell and m = 0 for pz orbital.
Hence electron will residue in 3pz orbital.
(ii) for s subshell. Hence electron will residue in 4s orbital.
Illustration 23. Write all the quantum numbers for the following orbitals.
(i) 3 (ii) 3s (iii) 4px
Solution: (i)
(ii)
(iii)
Pauli’s Exclusion Principle:
According to this principle, an orbital can contain a maximum number of two electrons and these two electrons must be of opposite spin.
Two electrons in an orbital can be represented by
↑↓ or ↓↑
SHAPES AND SIZE OF ORBITALS
An orbital is the region of space around the nucleus within which the probability of finding an electron of given energy is maximum (90–95%). The shape of this region (electron cloud) gives the shape of the orbital. It is basically determined by the azimuthal quantum number, while the orientation of orbital depends on the magnetic quantum number (m). Let us now see the shapes of orbitals in the various subshells.
s–orbitals: These orbitals are spherical and symmetrical about the nucleus. The probability of finding the electron is maximum near the nucleus and keep on decreasing as the distance from the nucleus increases. There is vacant space between two successive s–orbitals known as radial node. But there is no radial node for 1s orbital since it is starting from the nucleus. 
The size of the orbital depends upon the value of principal quantum number (n). Greater the value of n, larger is the size of the orbital. Therefore, 2s–orbital is larger than 1s orbital but both of them are nondirectional and spherically symmetrical in shape.
p–orbitals (=1):
The probability of finding the p–electron is maximum in two lobes on the opposite sides of the nucleus. This gives rise to a dumb–bell shape for the p–orbital. For p–orbital = 1. Hence,
m = –1, 0, +1. Thus, p–orbital have three different orientations. These are designated as px,py & pz depending upon whether the density of electron is maximum along the x y and z axis respectively. As they are not spherically symmetrical, they have directional character. The two lobes of p–orbitals are separated by a nodal plane, where the probability of finding electron is zero.
The three porbitals belonging to a particular energy shell have equal energies and are called degenerate orbitals.
d–orbitals (= 2):
For d–orbitals, l = 2. Hence m = –2,–1, 0, +1, +2. Thus there are 5d orbitals. They have relatively complex geometry. Out of the five orbitals, the three (dxy, dyz,dzx) project in between the axis and the other two and lie along the axis.
Illustration 24. An electron has a spin quantum number and a magnetic quantum
number −1. It cannot occupy a/an
(A) d orbital (B) f orbital
(C) p orbital (D) s orbital
Solution: For s orbital m = 0. Hence (D) is correct.
Illustration 25. Which dorbital has electron density in all the planes?
Solution: orbital has lobe along Zaxis and a belt of electrons in xy plane. Hence it has electron density in all xy, xz and yz plane.
RULES FOR FILLING OF ELECTRONS IN VARIOUS ORBITALS
The atom is built up by filling electrons in various orbitals according to the following rules.
Aufbau Principle: This principle states that the electrons are added one by one to the various orbitals in order of their increasing energy starting with the orbital of lowest energy. The increasing order of energy of various orbital is
1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,5f,6d,7p……………………
How to remember such a big sequence? To make it simple we are giving you the method to write the increasing order of the orbitals. Starting from the top, the direction of the arrow gives the order of filling of orbitals.
Alternatively, the order of increasing energies of the various orbitals can be calculated on the basis of (n +) rule.
The energy of an orbital depends upon the sum of values of the principal quantum number (n) and the azimuthal quantum number (). This is called (n +) rule. According to this rule,
“In neutral isolated atom, the lower the value of (n +) for an orbital, lower is its energy. However, if the two different types of orbitals have the same value of (n +), the orbitals with lower value of n has lower energy’’.
Illustration of (n +) Rule:
Type of orbitals  Value of n  Values of  Values of (n+)  Relative energy 
1s  1  0  1+0=1  Lowest energy 
2s  2  0  2+0=2  Higher energy than 1s orbital 
2p  2  1  2+1=3  2p orbital (n=2) have lower energy than 3s orbital (n=3) 
3s  3  0  3+1=4 
Hund’s Rule of Maximum Multiplicity:
This rule deals with the filling of electrons in the equal energy (degenerate) orbitals of the same sub shell (p,d and f). According to this rule,
“Electron pairing in p,d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied”.
This is due to the fact that electrons being identical in charge, repel each other when present in the same orbital. This repulsion can, however, be minimized if two electrons move as far apart as possible by occupying different degenerate orbitals. All the electrons in a degenerate set of orbitals will have same spin.
Electronic Configuration of Elements:
Electronic configuration is the distribution of electrons into different shells, subshells and orbitals of an atom.
Keeping in view the above mentioned rules, electronic configuration of any orbital can be simply represented by the notation]
Alternatively:
Orbital can be represented by a box and an electron with its direction of spin by arrow. To write the electronic configuration, just we need to know (i) the atomic number (ii) the order in which orbitals are to be filled (iii) maximum number of electrons in a shell, sub–shell or orbital.
 Each orbital can accommodate two electrons
 The number of electrons to be accomodated in a subshell is 2 × numbers of
degenerate orbitals.
Subshell  Maximum number of electrons 
s  2 
p  6 
d  10 
f  14 
 The maximum number of electrons in each shell (K, L, M, N…) is given by 2n2.
where n is the principal quantum number.  The maximum number of orbitals in a shell is given by n2 where n is the principal quantum number.
Importance of Knowing the Electronic Configuration:
The chemical properties of an element are dependent on the relative arrangement of its electrons.
Illustration 26. Write the electronic configuration of nitrogen (atomic number= 7)
Solution: 
Illustration 27. Write the electronic configuration of following:
(i)
Solution: (i)
(ii)
Exceptional Configurations:
Stability of Half Filled and Completely Filled Orbitals:
Cu has 29 electrons. Its expected electronic configuration is
1s22s22p63s23p64s23d9
But a shift of one electron from lower energy 4s orbital to higher energy 3d orbital will make the distribution of electron symmetrical and hence will impart more stability.
Thus the electronic configuration of Cu is
1s2 2s22p63s23p64s13d10
Fully filled and half filled orbitals are more stable.
Illustration 28. Which of the following metals has the highest value of exchange energy required for exchange stabilization?
(A) Mn (B) Cr
(C) Cu (D) Zn
Solution: Cu has full filled dsubshell. Hence (C) is correct.
Illustration 29. The total spin resulting from a d7 configuration is
(A) 1 (B) 2
(C) (D)
Solution: S = n/2, where n = number of unpaired electrons = 3. Hence (D) is correct.
Illustration 30. The electronic configuration of an element is 1s22s22p63s23p63d54s1. This represents its
(A) excited state (B) ground state
(C) cationic form (D) anionic form
Solution: It is ground state electronic configuration of Cr. Hence (B) is correct.
Illustration 31. Write the electronic configuration of the following ions:
(a) H+ (b) Na+ (c)O2− (d) F−
Solution: (a) 1s0 (b) 1s22s22p6 (c) 1s22s22p6 (d) 1s22s22p6
Illustration 32. We know the Hund’s rule. Explain how to arrange three electrons in p orbitals.
Solution: The important point to be remembered is, all the singly occupied orbitals should have parallel spins, i.e in the same direction eitherclockwise or anticlockwise.
2px 2py 2pz
↑  ↑  ↑ 
½ ½ ½ = 1½
The maximum multiplicity means that the total spin of unpaired electrons is maximum.
Illustration 33. We know that fully filled and half filled orbitals are more stable. Can you write the electronic configuration of Cr(Z = 24)?.
Solution: Cr (Z = 24)
1s2, 2s2,2p6,3s2,3p6,4s1,3d5.
Since half filled orbital is more stable one 4s electron is shifted to 3d orbital.
Illustration 34. The configuration of potassium (19) is 1s22s22p63s24s1 and not 1s22s22p63s23p63d1. Why?
Solution: According to Aufbau Principle’s electron first fill in a sublevel of lower energy level (lower n+). The energy of 4s(4 + 0 = 4) is lower than that of 3d (3 + 2 = 5), so the electronic configuration of K is 1s22s22p63s23p64s1.
Illustration 35. How many maximum number of electrons can be present in 3rd shell of an atom?
Solution: Total number of electrons in a shell = 2n2
= 2 × 32 = 18
Illustration 36. How many maximum number of electrons a dorbital can have?
Solution: Any orbital can have maximum of 2 electrons.
Illustration 37.Calculate the total number of delectrons in molybdenum. (atomic number = 42).
Solution: Electronic configuration of Mo is
Hence, total number of delectrons = 10 + 4 = 14
Exercise 4.
(i) Write the electronic configuration of the following:
(a) Cr3+ (b) Mn2+ (c) Cu
(ii) Write the orbital rotation for the following quantum numbers.
(a)
(b)
(c)
DUAL CHARACTER (PARTICLE AND WAVE CHARACTER OF MATTER AND RADIATION)
In case of light some phenomenon like diffraction and interference can be explained on the basis of its wave character. However, the certain other phenomenon such as black body radiation and photoelectric effect can be explained only on the basis of its particle nature. Thus, light is said to have a dual character. Such studies on light were made by Einstein in 1905.
Louis de Broglie, in 1924 extended the idea of photons to material particles such as electron and he proposed that matter also has a dual characteras wave and as particle.
Derivation of deBroglie Equation:
The wavelength of the wave associated with any material particle was calculated by analogy with photon.
In case of photon, if it is assumed to have wave character, its energy is given by
E = hν …(i) (according to the Planck’s quantum theory)
where ν is the frequency of the wave and ‘h’ is Planck’s constant
If the photon is supposed to have particle character, its energy is given by
E = mc2 … (ii) (according to Einstein’s equation)
where ‘m’ is the mass of photon, ‘c’ is the velocity of light.
By equating (i) and (ii)
hν = mc2
But ν = c/λ
h = mc2
(or) λ = h /mc
The above equation is applicable to material particle if the mass and velocity of photon is replaced by the mass and velocity of material particle. Thus for any material particle like electron.
λ = h/mv or λ = where mv = p is the momentum of the particle.
Relation Between Kinetic Energy and Wavelength:
K.E (E)=1/2 mv2
v = ⇒
Derivation of Angular Momentum from de Broglie Equation:
According to Bohr’s model, the electron revolves around the nucleus in circular orbits. According to de Broglie concept, the electron is not only a particle but has a wave character also.
If the wave is completely in phase, the circumference of the orbit must be equal to an integral multiple of wave length (λ)
Therefore 2πr = nλ where ‘n’ is an integer and ‘r’ is the radius of the orbit But λ = h/mv ∴ 2πr = nh /mv or mvr = n h/2π 
which is Bohr’s postulate of angular momentum, where ‘n’ is the principal quantum number.
“Thus, the number of waves an electron makes in a particular Bohr orbit in one complete revolution is equal to the principal quantum number of the orbit”.
Alternatively:
Number of waves ‘n’ = = =
Where v and r are the velocity of electron and radius of that particular Bohr orbit in which number of waves are to be calculated, respectively.
The electron is revolving around the nucleus in a circular orbit. How many revolutions it can make in one second?
Let the velocity of electron be v m/sec. The distance it has to travel for one revolution 2πr, (i.e., the circumference of the circle).
Thus, the number of revolutions per second is =
Common unit of energy is electron volt which is amount of energy given when an electron is accelerated by a potential of exactly 1 volt. This energy equals the product of voltage and charge. Since in SI units coulombs × volts = joules, 1 eV numerically equals the electronic charge except that joule replaces coulombs.
Illustration 38. An e–, a proton and an alpha particle have K.E of 16 E, 4 E and E respectively. What’s the qualitative order of their Broglie wavelengths?
(A) λe > λP > λa (B) λe > λp = λa
(C) λP < λe < λa (D) λa < λe = λP
Solution: λ=. Hence (B) is correct.
Illustration 39. Calculate the de Broglie wavelength of a ball of mass 0.1 kg moving with a speed of 60 ms–1.
Solution: =
λ = 1.1 × 10–34 m.
This is apparent that this wavelength is too small for ordinary observation.
Although the de Broglie equation is applicable to all material objects but it has significance only in case of microscopic particles.
Since, we come across macroscopic objects in our everyday life, de Broglie relationship has no significance in everyday life.
[Distinction between the wave particle nature of a photon and the particle wave nature of a sub atomic particle]:
Photon  Sub Atomic Particle 
1. Energy = hν  Energy = mv2 
2. Wavelength =  Wavelength = 
Note: We should never interchange any of the above
Illustration 40. Calculate the number of waves made by electron in 4th Bohr’s orbit.
Solution: Number of waves made by electron in nth orbit = n
Hence the number of waves in 4rd orbit = 4
Illustration 41. What is the deBroglie wavelength of electron having K.E. of 5 eV?
Solution: K.E. =
=
=
= 5.486 × 10−10 m
Illustration 42. Through what potential difference an electron must be accelerated to have a
deBroglie wavelength of 1.
Solution: λ = 10−10 m
We know K.E. =
v =
Now
Hence
= 1.5 × 106 volt
HEISENBERG’S UNCERTAINTY PRINCIPLE
All moving objects that we see around us e.g., a car, a ball thrown in the air etc., move along definite paths. Hence their position and velocity can be measured accurately at any instant of time. Is it possible for subatomic particle also?
As a consequence of dual nature of matter, Heisenberg, in 1927 gave a principle about the uncertainties in simultaneous measurement of position and momentum (mass × velocity) of small particles.
This Principle States:
“It is impossible to measure simultaneously the position and momentum of a small microscopic moving particle with absolute accuracy or certainty” i.e., if an attempt is made to measure any one of these two quantities with higher accuracy, the other becomes less accurate.
The product of the uncertainty in position (Δx) and the uncertainty in the momentum (Δp = m.Δv where m is the mass of the particle and Δv is the uncertainty in velocity) is equal to or greater than h/4π where h is the Planck’s constant.
Thus, the mathematical expression for the Heisenberg’s uncertainty principle is simply written as
Δx.Δp ≥ h/4π
Explanation of Heisenberg’s uncertainty principle
Suppose we attempt to measure both the position and momentum of an electron, to pinpoint the position of the electron we have to use light so that the photon of light strikes the electron and the reflected photon is seen in the microscope. As a result of the hitting, the position as well as the velocity of the electron are disturbed. The accuracy with which the position of the particle can be measured depends upon the wavelength of the light used. The uncertainty in position is ±λ. The shorter the wavelength, the greater is the accuracy. But shorter wavelength means higher frequency and hence higher energy. This high energy photon on striking the electron changes its speed as well as direction. But this is not true for macroscopic moving particle. Hence Heisenberg’s uncertainty principle is not applicable to macroscopic particles.
Illustration 43. Why electron cannot exist inside the nucleus according to Heisenberg’s uncertainty principle?
Solution: Diameter of the atomic nucleus is of the order of 10–15m
The maximum uncertainty in the position of electron is 10–15 m.
Mass of electron = 9.1 ×10–31 kg.
Δx. Δp =
Δx × (m.Δv) = h/4π
Δv = = ×
Δv = 5.80 × 1010 ms–1
This value is much higher than the velocity of light and
hence not possible.
Illustration 44. What is the uncertainity in the position of electron, if uncertainity in its velocity is 0.0058 m/s?
Solution:
Δx = 0.01 m
Illustration 45. What is the uncertainity in the position of a ball of mass 10 g and which is moving with a velocity of 100 m/s with 0.002 % uncertainity?
Solution:
Now
Exercise 5.
(i) Two particles A and B are in motion. If the wavelength associated with particle A is 5 ×10–8 m, calculate the wavelength associated with particle B if its momentum is half of A.
(ii) Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 kV.
(iii) Calculate the uncertainty in position of a particle when uncertainty in the momentum is
(a) 1×10–2 gm cm sec–1 and (b) zero.
QUANTUM MECHANICAL MODEL OF ATOM
The atomic model which is based on the particle and wave nature of the electron is known as wave or quantum mechanical model of the atom. This was developed by Erwin Schrodinger in 1926. This model describes the electron as a three dimensional wave in the electronic field of positively charged nucleus. Schrodinger derived an equation which describes wave motion of an electron. The differential equation is
where x, y, z are certain coordinates of the electron, m = mass of the electron E = total energy of the electron. V = potential energy of the electron; h = Planck’s constant and ψ (psi) = wave function of the electron.
Significance of ψ: The wave function may be regarded as the amplitude function expressed in terms of coordinates x, y and z. The wave function may have positive or negative values depending upon the value of coordinates. The main aim of Schrodinger equation is to give solution for probability approach. When the equation is solved, it is observed that for some regions of space the value of ψ is negative. But the probability must be always positive and cannot be negative, it is thus, proper to use ψ2 in favour of ψ.
Significance of ψ2: ψ2 is a probability factor. It describes the probability of finding an electron within a small space. The space in which there is maximum probability of finding an electron is termed as orbital. The important point of the solution of the wave equation is that it provides a set of numbers called quantum numbers which describe energies of the electron in atoms, information about the shapes and orientations of the most probable distribution of electrons around nucleus.
Nodal Points and Planes:
The point where there is zero probability of finding the electron is called nodal point. There are two types of nodes: Radial nodes and angular nodes. The former is concerned with distance from the nucleus while latter is concerned with direction.
No. of radial nodes = n – – 1
No. of angular nodes =
Total number of nodes = n – 1
Nodal planes are the planes of zero probability of finding the electron. The number of such planes is also equal to .
Illustration 46. Calculate radial nodes and angular nodes for the following type of orbitals.
(a) 1s, (b) 2p, (c) 3p, (d) 3d, (e) 4s and (f) 4d
Solution: (a) 0, 0 (b) 0, 1
(c) 1, 1 (d) 0, 2
(e) 3, 0 (f) 1, 2
PHOTOELECTRIC EFFECT
Sir J.J. Thomson, observed that when a light of certain frequency strikes the surface of a metal, electrons are ejected from the metal. This phenomenon is known as photoelectric effect and the ejected electrons are called photoelectrons.
A few metals, which are having low ionization energy like Cesium, show this effect under the action of visible light but many more show it under the action of more energetic ultraviolet light.
An evacuated tube contains two electrodes connected to a source of variable voltage, with the metal plate whose surface is irradiated as the anode. Some of the photoelectrons that emerge from this surface have enough energy to reach the cathode despite its negative polarity, and they constitute the measured current. The slower photoelectrons are repelled before they get to the cathode. When the voltage is increased to a certain value V0, of the order of several volts, no more photoelectrons arrive, as indicated by the current dropping to zero. This extinction voltage (or also referred as stopping potential) corresponds to the maximum photoelectron kinetic energy i.e., eVo = ½ mv2
The experimental findings are summarised as below:
 Electrons come out as soon as the light (of sufficient energy) strikes
the metal surface.  The light of any frequency will not be able to cause ejection of electrons from a metal surface. There is a minimum frequency, called the threshold (or critical) frequency, which can just cause the ejection. This frequency varies with the nature of the metal. The higher the frequency of the light, the more energy the photoelectrons have. Blue light results in faster electrons than red light.
 Photoelectric current is increased with increase in intensity of light of same frequency, if emission is permitted i.e., a bright light yields more photoelectrons than a dim one of the same frequency, but the electron energies remain the same.
Light must have stream of energy particles or quanta of energy (hν). Suppose, the threshold frequency of light required to eject electrons from a metal is ν0, when a photon of light of this frequency strikes a metal it imparts its entire energy (hν0) to the electron.
“This energy enables the electron to break away from the atom by overcoming the attractive influence of the nucleus”. Thus each photon can eject one electron. If the frequency of light is less than ν0 there is no ejection of electron. If the frequency of light is higher than ν0 (let it be ν), the photon of this light having higher energy (hν), will impart some energy to the electron that is needed to remove it from the atom. The excess energy would give a certain velocity (i.e, kinetic energy) to the electron.
hν = hν0 + K.E
hν = hν0 + ½ mv2
½ mv2 = hν–hνo
where, ν = frequency of the incident light
ν0 = threshold frequency
hν0 is the threshold energy (or) the work function denoted by φ = hν0 (minimum energy of the photon to liberate electron). It is constant for particular metal and is also equal to the ionization potential of gaseous atoms.
The kinetic energy of the photoelectrons increases linearly with the frequency of incident light. Thus, if the energy of the ejected electrons is plotted as a function of frequency, it result in a straight line whose slope is equal to Planck’s constant ‘h’ and whose intercept is hν0.
Illustration 47. Work function of sodium is 2.5 eV. Predict whether the wavelength 6500 is suitable for a photoelectron ejection or not.
Solution: Energy of incident light
= 3.055 × 10−19 J
= 1.9 eV
Which is lower than work function. Hence no ejection will take place.
Exercise 6.
(i) Find the threshold wavelengths for photoelectric effect from a copper surface, a sodium surface and a caesium surface. The work function of these metal are
4.5 eV, 2.3 eV and 1.9 eV respectively.
(ii) Energy required to stop the ejection of electrons from Cu plate is 0.24 eV.
Calculate the work function when radiation of λ = 253.7 nm strikes the plate?
ANSWER TO EXERCISES
Exercise 1:
(i) 19
(ii) (a) 1.2044 × 1022
(b) 2.01 × 10–5 kg
Exercise2:
(i)
(ii)
Exercise 3:
(i) 6603 Å
(ii) 9.84 × 105 J mol–1
(iii) 486 nm, Blue
Exercise 4:
(i) (a)
(b)
(c)
(ii) (a) 4d
(b) 2p
(c) 3s
Exercise 5:
(i) 10–7 m
(ii) 3.87 × 10–11 m
(iii) (a) 5.27 × 10–28m
(b) ∞
Exercise 6:
(i) 276 nm, 540 nm, 654 nm
(ii) 4.65 eV
MISCELLANEOUS EXERCISES
Exercise 1: Calculate the ratio of specific charge (e/m) of a proton and that of an αparticle.
Exercise 2: What is the fraction of volume occupied by the nucleus with respect to the total volume of an atom?
Exercise 3: Calculate the minimum and maximum values of wavelength in Balmer series of a H atom.
Exercise 4: The wave number of the first line of Balmer series of hydrogen is 15,200 cm−1. What is the wave number of the first line of Balmer series for the Li2+ ion?
Exercise 5: What possible can be the ratio of the de Broglie wavelengths for two electrons having the same initial energy and accelerated through 50 volts and 200 volts?
Exercise 6: What is likely to be the principle quantum number for a circular orbit of diameter 20 nm of the hydrogen atom if we assume Bohr orbit to be the same as that represented by the principle quantum number?
Exercise 7: Calculate the uncertainty in the velocity of a particle weighing 25.0 g if the uncertainty in position is 10−5 m. (given Planck constant h = 6.6 × 10−34 Js).
Exercise 8: Calculate the de Broglie wavelength of a tennis ball of mass 60.0 g moving with a velocity of 10 metres per second (Planck constant, h = 6.63 × 10−34 Js).
Exercise 9: Suppose 10–17J of energy is needed by the interior of human eye to see an object. How many photons of green light (λ=550 nm) are needed to generate this minimum amount of energy?
Exercise 10: Ionization potential of hydrogen atom 13.6 eV. If hydrogen atom in ground state excited by monochromatic light of energy 12.1 eV, then what will be the total spectral lines emitted according to Bohr’s theory?
ANSWERS TO MISCELLANEOUS EXERCISES
Exercise 1: 1 : 2
Exercise 2: 10−15
Exercise 3: 3647 nm and 6564 nm
Exercise 4: 1,36,800 cm−1
Exercise 5: 2 : 1
Exercise 6: 14
Exercise 7: 2.1 × 10−28
Exercise 8: 10−33 m
Exercise 9: 28
Exercise 10: 3
SOLVED PROBLEMS
Subjective:
Board Type Questions
Prob 1. Why are Bohr’s orbits are called stationary states?
Sol. This is because the energies of orbits in which the electrons revolve are fixed.
Prob 2. Explain why the electronic configuration of Cu is 3d104s1 and not 3d94s2.
Sol. In the 3d104s1 the dsubshell is completely filled which is more stable.
Prob 3. Fe3+ ion is more stable than Fe2+ ion. Why?
Sol. In Fe3+ ion 3d subshell is half filled hence more stable configuration.
Prob 4. Calculate the accelerating potential that must be applied to a proton beam to give it an effective wavelength of 0.005 nm.
Sol.
Putting the values we get
V = 32.85 volt
Prob 5. Give one example of isodiapheres.
Sol. Isodiapheres have same difference between the number of neutrons and protons. For example
IIT Level Questions
Prob 6. Which electronic transition in Balmer series of hydrogen atom has same frequency as that of n = 6 to n = 4 transition in He+. [Neglect reduced mass effect].
Sol.
= =
= R × 12
=
On solving above equation
n2 = 9
∴ n = 3
Or corresponding transition from 3 → 2 in Balmer series of hydrogen atom has same frequency as that of 6 → 4 transition in He+.
Prob 7. Calculate ionization potential in volts of (a) He+ and (b) Li2+
Sol. I.E. =
= [Z =2 for He+]
= 13.6 × 4 = 54.4 eV
Similarly for Li2+ =
= 13.6 × 9 = 122.4 eV
Prob 8. Calculate the ratio of K.E and P.E of an electron in an orbit?
Sol. K.E. =
P.E. =
∴ P.E. = –2K.E
∴
Prob 9. How many spectral lines are emitted by atomic hydrogen excited to
nth energy level?
Sol. 
Thus the number of lines emitted from nth energy level
= 1+2+3+ …………. n–1 = Σ (n–1)
Σn =
∴ Σ(n–1)= =
Number of spectral lines that appear in hydrogen spectrum when an electron jumps from nth energy level =
Prob 10. Calculate (a) the de Broglie wavelength of an electron moving with a velocity of
5.0 × 105 ms–1 and (b) relative de Broglie wavelength of an atom of hydrogen and atom of oxygen moving with the same velocity (h = 6.63 × 10–34 kg m2 s–1)
Sol. (a) λ = =
Wavelength λ = 1.46 × 10–9m
(b) An atom of oxygen has approximately 16 times the mass of an atom of hydrogen. In the formula, h is constant while the conditions of problem make v, also constant. This means that λ and m are variables and λ varies inversely with m. Therefore, λ for the hydrogen atom would be 16 times greater than λ for oxygen atom.
Prob 11. A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest approach for head on collision.
Sol. K.E. = P.E.
Hence d =
Prob 12. What is the wavelength associated with 150 eV electron
Sol. λ =
= = = 10–10 m = 1 Å
Prob 13. The energy of electron in the second and third Bohr orbit of the hydrogen atom is
–5.42 × 10–12 erg and –2.41 × 10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.
Sol. E3 – E2 = hν =
– 2.41 × 10–12 – (– 5.42 × 10–12) =
∴ λ =
= 6.604 × 10–5 cm = 6.604 × 10–5 ×108 = 6604 Å
Prob 14. O2 undergoes photochemical dissociation into one normal oxygen and one excited oxygen atom, 1.967 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen atoms requires 498 kJ mole–1. What is the maximum wavelength effective for photochemical dissociation of O2?
Sol. O2 ⎯→ ON +
O2 ⎯→ ON + ON
E = 498 ×103 J / mole
= per molecule = 8.268 ×10–19 J
Energy required for excitation = 1.967 eV = 3.146 ×10–19J
Total energy required for photochemical dissociation of O2
= 8.268 ×10–19 + 3.146 ×10–19 = 11.414 × 10–19 J = 11.414 ×10–19 J
λ = = 1.7415 ×10–7 m = 1741.5 Å
Prob 15. Compare the wavelengths for the first three lines in the Balmer series with those which arise from similar transition in Be3+ ion. (Neglect reduced mass effect).
Sol.
∴ = 16
So we can conclude that all transitions in Be3+ will occur at wavelengths times the hydrogen wavelengths.
Objective:
Prob 1. For a pelectron, orbital angular moment is
(A) (B)
(C) (D)
Sol. Orbital angular momentum L = where
∴ L for p electron =
∴ (A)
Prob 2. For which of the following species, Bohr theory doesn’t apply
(A) H (B) He+
(C) Li2+ (D) Na+
Sol. Bohr theory is not applicable to multi electron species
∴ (D)
Prob 3. If the radius of 2nd Bohr orbit of hydrogen atom is r2. The radius of third Bohr orbit will be
(A) (B) 4r2
(C) (D) 9r2
Sol. ∴
∴ r3 =
∴ (C)
Prob 4. Number of waves made by an electron in one complete revolution in 3rd Bohr orbit is
(A) 2 (B) 3
(C) 4 (D) 1
Sol. Circumference of 3rd orbit = 2πr3
According to Bohr’s angular momentum of electron in 3rd orbit is
mvr3 = or
By deBroglie equation,
λ =
∴ λ =
∴ 2πr3 = 3λ
i.e. circumference of 3rd orbit is three times the wavelength of electron or number of waves made by Bohr electron in one complete revolution in 3rd orbit is three.
∴ (B)
Prob 5. The degeneracy of the level of hydrogen atom that has energy is
(A) 16 (B) 4
(C) 2 (D) 1
Sol. En =
∴
i.e. for 4th subshell
i.e. 1 + 3 + 5 + 7 = 16
∴ Degeneracy is 16
Prob 6. An electron is moving with a kinetic energy of 4.55 ×10–25 J. What will be
de Broglie wavelength for this electron?
(A) 5.28 ×10–7 m (B) 7.28 ×10–7 m
(C) 2 ×10–10 m (D) 3 ×10–5 m
Sol. KE = mv2 = 4.55 × 10–25
v2 =
v = 103 m/s
de Broglie wavelength λ= = = 7.28 ×10–7 m
∴ (B)
Prob 7. Suppose 10–17J of energy is needed by the interior of human eye to see an object. How many photons of green light (λ = 550 nm) are needed to generate this minimum amount of energy?
(A) 14 (B) 28
(C) 39 (D) 42
Sol. Let the number of photons required = n
n
n = = = 27.6 = 28 photons
∴ (B)
Prob 8. The two electrons present in an orbital are distinguished by
(A) principal quantum number (B) azimuthal quantum number
(C) magnetic quantum number (D) spin quantum number
Sol. They are distinguished by their spin.
∴ (D)
Prob 9. The velocity of electron in the ground state hydrogen atom is 2.18 × 106 ms–1. Its velocity in the second orbit would be
(A) 1.09 ×106 ms–1 (B) 4.38 ×106 ms–1
(C) 5.5 ×105 ms–1 (D) 8.76 ×106 ms–1
Sol. We know that velocity of electron in nth Bohr’s orbit is given by
ν = 2.18 × 106 m/s
For H, Z = 1
v1 = m/s
v2 = m/s = 1.09 ×106 m/s
∴(A)
Prob 10. The ionization energy of the ground state hydrogen atom is 2.18×10–18J. The energy of an electron in its second orbit would be
(A)–1.09 ×10–18 J (B) –2.18 ×10–18J
(C) –4.36 ×10–18J (D) –5.45 ×10–19J
Sol. Energy of electron in first Bohr’s orbit of H–atom
E = ( ionization energy of H = 2.18 ×10–18J)
E2 = J = –5.45 ×10–19J
∴(D)
True and False
Prob 11. All electromagnetic radiations have same energy.
Sol. False. Different electromagnetic radiation have different wavelengths, hence different energy
Prob 12. Bohr’s model is applicable for Hatom only.
Sol. False. Bohr’s model is applicable for species having one electron like H, He+, Li2+.
Prob 13. Specific charge of cathode rays remains same irrespective of the gas used in the discharge tube.
Sol. True. Cathode rays contain electrons which remains same with every gas.
Prob 14. Isoelectronic species have same electronic configuration.
Sol. True. Isoelectronic species have same number of electrons.
Prob 15. Atoms with same atomic number but different number of neutrons are called isotopes.
Sol. True. Isotopes have same atomic number but different atomic mass.
Fill in the Blanks
Prob 16. The mass of positron is ___________ electron.
Sol. Equal to
Prob 17. With increasing principle quantum number, the energy difference between adjacent energy levels in Hatom _________.
Sol. Decreases
Prob 18. If the magnetic moment of an ion An+ is 5.9, then the number of unpaired electrons present in An+ are ____________.
Sol. Five. Applying μ =
Prob 19. selectrons are more penetrating than pelectrons. The energy required to abstract
selectrons is ____________ than for p electrons.
Sol. Greater
Prob 20. The wavelength of the first spectral line in Paschen series for Hatom is____________.
Sol. 16410 Å
For Paschen series
ASSIGNMENT PROBLEMS
Subjective:
Level – O
 How many electrons in an atom have the following quantum numbers?
(i) n = 4, ms = −1/2 (ii) n = 3,
 Write the number of protons, electrons and neutrons in nitride ion.
 Distinguish between absorption and emission spectra.
 Write down the electronic configuration of Mn4+ and Cr3+. How many unpaired electrons are present in them?
 Write correct orbital notation for each of the following sets of quantum numbers.
(i) n = 1, , m = 0 (ii) n = 2, , m = −1
 What are drawbacks of Rutherford’s model atom?
 Explain why uncertainty principle is significant only for the motion of subatomic particles but not for macroscopic objects.
 Why we do not see a car moving in the form of a wave on a road?
 Why e/m ratio in case of anode rays depends upon the type of gas?
 Why splitting of spectral lines take place when the source giving the spectrum is placed in a magnetic field?
 What is the maximum number of lines obtained when the excited electrons of a hydrogen atom in n = 6 drops to the ground state.
 Calculate the radius of the third orbit of a hydrogen atom.
 The wavelength of a moving body of mass 0.1 mg is Calculate its kinetic energy
 Calculate the mass and charge of one mole of electrons.
 Calculate (a) wave number and (b) frequency of yellow radiations having wavelength of 5800.
 In one discharge tube H2 gas is taken and in other O2 gas is taken. Will the negative and positive charged particles in cathode rays and anode rays be same?
 Explain why half filled and full filled configurations are more stable.
 Some energy is absorbed by hydrogen atom due to which an electron in it jumped from ground state to the state having principle quantum number 5 and again jumped back to the original level. What type of spectrum is observed and in which region. What is the name of series?
Level – I
 What is the mass of a photon of sodium light with a wavelength of 5890?
 Calculate the kinetic energy of a moving electron which has a wavelength of 4.8 pm (Mass of electron
 Calculate de – Broglie wavelength of an electron moving with 1% of the speed of light.
 Calculate the energy of an electron in the second Bohr orbit of an excited hydrogen atom
 Calculate the velocity of an electron revolving in the second orbit of a hydrogen atom
 How many photons of light having a wavelength of 4000 are necessary to provide 1 J of energy? (h = 6.63×1034Js, c = 3×108 m/s)
 How the electron in the ground state of hydrogen atom is excited by means of monochromatic radiation of wavelength 970.6. How many different lines are possible in the resulting emission spectrum? Find the longest wavelength among these?
 The wavelength of the first line in the Balmer series is 656 nm. Calculate the wavelength of the second line and the limiting line in Balmer series.
 A single electron system has ionization energy 11180 kJ / mol. Find the number of protons in the nucleus of the system.
 Calculate the velocity of an electron placed in the third orbit of hydrogen atom. Also calculate the number of revolutions per second that this electron makes around the nucleus.
 (a) What is wavelength of a particle of mass 1 g moving with a velocity of 200 m/s?
(b) A moving electron has 4.55 ×1025 joules of kinetic energy. Calculate its wavelength (mass = 9.1 ×10–31 kg and h = 6.6×10–34 kg m2sec–1)
(c) Calculate the deBroglie wavelength of electron accelerated through 100 volt.
 Calculate the radii of the I, II and III permitted electron Bohr orbits in a hydrogen atom. What are the corresponding values in the case of a singly ionised helium atom?
 Calculate energy in kcal/mole necessary to remove an electron in a hydrogen atom in fourth principal quantum number to infinity.
 A monochromatic source of light operating at 600 watt emits 2 ×1022 photons per second. Find the wavelength of the light.
 Calculate the uncertainty in the position of an electron if it has a speed of 500 m/s with an uncertainity of 0.02%.
Level – II
 (a) The electron energy in hydrogen atom is given by E = – 21.7 × 10–12/n2 erg. Calculate the energy required to remove an electron completely from n = 2 orbit. What is the longest wavelength (in cm) of light that can be used to cause this transition?
(b) Ionization energy of hydrogen atom is 13.6 eV. Calculate the ionization energy for Li2+ and Be3+ in the first excited state.
 A bulb emits light of wave length 4500 Å. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by bulb per second?
 (a) What Amount of accelerating potential is needed to produce an electron beam with an effective wavelength of 0.09 Å.
(b) Calculate deBroglie wavelength of an electron moving with a speed of nearly th that of light (3 × 108 ms–1)
 A doubly ionised lithium atom is hydrogen like with an atomic no. 3.
(a) Find the wavelength of the radiation required to excite the electron in Li from the first to the third Bohr orbit (ionisation energy of the hydrogen atom is equal to 13.6 eV)
(b) How many spectral lines are observed in the emission spectrum of the above excited system?
 Calculate λ of the radiations when the electron jumps from III to II orbit of hydrogen atom. The electronic energy in II and III Bohr orbit of hydrogen atoms are
– 5.42 ×10–12 and –2.41 ×10–12erg respectively.
 1.8 g hydrogen atoms are excited to radiations. The study of spectra indicates that 27% of the atoms are in 3rd energy level and 15% of atoms in 2nd energy level and the rest in ground state. IP of H is 13.6 eV. Calculate (i) No. of atoms present in III and II energy level (ii) Total energy evolved when all the atoms return to ground state.
 The IP of H is 13.6 eV. It is exposed to electromagnetic waves of 1028 Å and given out induced radiations. Find the wavelength of these induced radiations.
 What is uncertainty in velocity of an electron if uncertainty in its position is 1Å?
 A single electron beam, atom has nuclear charge +Ze where Z is atomic number and e is electronic charge. It requires 16.52 eV to excite the electron from the second Bohr orbit to third Bohr orbit. Find
(a) The atomic no. of element
(b) The energy required for transition of electron from first to third orbit.
(c) Wavelength required to remove electron from first Bohr orbit to infinity.
(d) The kinetic energy of electron in first Bohr orbit.
 A sample of hydrogen gas containing some atoms in one excited state emitted three different types of photons. When the sample was exposed with radiation of energy
2.88 eV it emitted 10 different types of photons, all having energy equal or less than 13.05 eV. Find out
(a) The principal quantum numbers of initially excited electrons.
(b) The principal quantum numbers of electrons in final excited state.
(c) The maximum and minimum energies of initially emitted photons.
Objective:
Level – I
 The ratio of energy of the electron in ground state of the hydrogen to electron in first excited state of He+ is
(A) 1 : 4 (B) 1 : 1
(C) 1 : 8 (D) 1 : 16
 Bohr model can explain spectrum of
(A) the hydrogen atom only
(B) all elements
(C) any atomic or ionic species having one electron only
(D) the hydrogen molecule
 Which is the correct order of probability of being found close to the nucleus is
(A) s > p > d > f (B) f > d > p > s
(C) p > d > f > s (D) d > f > p > s
 Ratio between longest wavelengths of H atom in Lyman series to the shortest wavelength in Balmer series of He+is
(A) (B)
(C) (D)
 If the radius of first Bohr orbit is a, then deBroglie wavelength of electron in 3rd orbit is nearly.
(A) 2πa1 (B) 6πa1
(C) 9πa1 (D) 16πa1
 If uncertainty in position and momentum are equal, the uncertainty in velocity would be
(A) (B)
(C) (D)
 Which of the following is a coloured ion?
(A) Cu+(aq) (B) Na+(aq)
(C) Cu2+ (aq) (D) K+ (aq)
 The number of orbitals in a subshell are given by
(A) 2 (B) n2
(C) 2 + 1 (D) 2n2
 For a ‘d’ electron, the orbital angular momentum is
(A) (B)
(C) (D)
 Hydrogen atom consists of a single electron but so many lines appear in the spectrum of atomic hydrogen because
(A) Sample contains some impurity
(B) Experiment is done on collection of atoms.
(C) Hydrogen atom splits to form more than one different species.
(D) Some different isotope of hydrogen atom may be present.
 A line with wave number 1.028×102 nm1 is emitted in the spectrum of atomic hydrogen. In what region of the electromagnetic spectrum does this line occur?
(A) far UV (B) near UV
(C) near IR (D) far IR
 One molecule of a substance absorbs one quantum of energy. The energy involved when 1.5 mole of the substance absorbs red light of frequency 7.5 ×1014 sec–1 will be
(A) 2.99 ×105 J (B) 3.23 ×105J
(C) 4.48 ×105J (D) 2.99 ×106 J
 If a shell is having g subshell, which is correct statement about principal quantum number n of this shell.
(A) n ≤ 5 (B) n ≥ 5
(C) n = 5 (D) Cannot be determined
 The wave number of a spectral line is 5 ×105 m–1. The energy corresponding to this line will be
(A) 3.39×10–23 kJ (B) 9.93 ×10–23 kJ
(C) 3.45 ×10–24 J (D) none of these
 Wave length of the radiation when electron jumps from second shell to 1st shell of H atom
(RH = 109679 cm–1)
(A) 1215.6 Å (B) 1397.5 Å
(C) 2395.87 Å (D) none of these
 The orbital which have lobe along the axis is/are
(A) (B) dxz
(C) (D) dxy
 Anode rays
(A) contain positively charged particles
(B) have constant e/m ratio irrespective of the type of gas used
(C) may contain He+ ions
(D) are always beams of protons
 The spectrum of He is expected to be similar to that of
(A) H (B) Li+
(C) Be2+ (D) H+
 The total energy of electron in an atom
(A) is less than zero
(B) is the sum of kinetic energy and potential energy
(C) is equal to kinetic energy in magnitude
(D) is equal to potential energy in magnitude
 Which of the following orbitals have no nodal plane?
(A) px (B) dxy
(C) (D) 4s
Assertion–Reason type questions
The following questions consist of two statements each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following five responses.
(A) If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.
(B) If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.
(C) If Assertion is true but the Reason is false.
(D) If the Assertion is false but the Reason is true.
(E) If both Assertion and Reason are false.
 A: A spectral line will be observed for 2px2py transition.
R: The energy will be released in the form of electromagnetic radiations.
 A: Cs is used in photoelectric cells.
R: Cs is an alkali metal.
 A: The energy of an electron is largely determined by its principle quantum number.
R: 2s subshell has higher energy than 1s.
 A: Fe3+(g) ion is more stable than Fe2+(g).
R: Fe3+(g) ion has higher magnetic moment than Fe2+(g) ion.
 A: deBroglie’s equation and uncertainity principle are applicable to moving bodies.
R: These have siginificance to microscopic as well as macroscopic particles.
Level – II
 Which statement is wrong about Bohr’s theory?
(A) Orbit is a three dimensional area where probability of finding electron is maximum.
(B) Orbit is a two dimensional track on which electron moves
(C) Atom has definite boundary
(D) Energies and angular momentum of orbits are quantized.
 The introduction of a neutron into the nuclear composition of an atom would lead to a change in
(A) The number of electrons also (B) The chemical nature of the atom
(C) Its atomic number (D) Its atomic weight
 Which statement is true?
(A) Spacing between energy levels n = 1 and n = 2 in hydrogen atom is greater than that of
n = 2 and n = 3
(B) Spacing between energy levels n = 1 and n = 2 in hydrogen atom is equal to that n = 2 and n = 3
(C) Spacing between energy levels n = 1 and n = 3 in hydrogen atom is less than that of
n = 2 and n = 3
(D) None of these
 If the wavelength of first line of the Balmer series of hydrogen atom is 656.1 nm, the wavelength of second line of this series would be
(A) 218.7 nm (B) 328.0 nm
(C) 486.0 nm (D) 640.0 nm
 In absence of external magnetic field f subshell is
(A) 5 fold degenerate (B) 3 fold degenerate
(C) 7 fold degenerate (D) nondegenerate
 Which set of quantum numbers is not possible for electron in 3rd shell?
(A) n = 3 = 2 m = – 1 s = +1/2
(B) n = 3 = 2 m = – 1 s = –1/2
(C) n = 3 = 2 m = 0 s = +1/2
(D) n = 3 = 3 m = 0 s = –1/2
 The first four ionization energies of an element are 191, 578,872 and 5962 kcal. The number of valence electrons in the element is
(A) 1 (B) 2
(C) 3 (D) 4
 The radiation is emitted when a hydrogen atom goes from a high energy state to a lower energy state. The wavelength of one line in visible region of atomic spectrum of hydrogen is 6.5 × 10–9 m. Energy difference between the two states is
(A) 3.0 ×10–17 J (B) 1.0 ×10–18 J
(C) 5.0 × 10–10 J (D) 6.5 ×10–7 J
 The ratio of the energy of the electron in ground state of hydrogen to the electron in first excited state of Be3+ is
(A) 1 : 4 (B) 1 : 8
(C) 1 : 16 (D) 16 : 1
 Which one of the following species is isoelectronic with P3–?
(A) Kr (B) Ca2+
(C) Na+ (D) F–
 Which of the following sets of quantum numbers is/are not allowable?
(A) n = 3 = 2 m = 0
(B) n = 2 = 0 m = – 1
(C) n = 4 = 3 m = + 1
(D) n = 1 = 0 m = 0
 Among V (Z = 23), Cr (Z = 24), Mn (Z = 25) which will have highest magnetic moment.
(A) V
(B) Cr
(C) Mn
(D) all of them will have equal magnetic moment
 If velocity of an electron in 1st Bohr orbit of hydrogen atom is x, its velocity in 3rd orbit will be
(A) (B) 3x
(C) 9x (D)
 Which element has a hydrogen like spectrum whose lines have wavelengths one fourth of atomic hydrogen?
(A) He+ (B) Li 2+
(C) Be3+ (D) B4+
 Calculate the wavelength of a track star running 150 metre dash in 12.1 sec if its weight is
50 kg.
(A) 9.11 × 10–34 m (B) 8.92 × 10–37 m
(C) 1.12 × 10–45 metre (D) none of these
16. Which of the following sets of quantum numbers is not allowed
(A) n = 3, = 1, m = +2 (B) n = 3, = 1, m = +1
(C) n = 3, = 0, m = 0 (D) n = 3, = 2, m = ± 2
 Assuming that a 25 watt bulb emits monochromatic yellow light of wavelength 0.57μ. The rate of emission of quanta per sec. will be
(A) 5.89 × 1015 sec–1 (B) 7.28 × 1017 sec–1
(C) 5 × 1010 sec–1 (D) 7.18 × 1019 sec–1
 How many chlorine atoms can you ionize in the process Cl ⎯→ + e, by the energy liberated from the following process?
Cl + e– ⎯→ for 6 × 1023 atoms
Given electron affinity of Cl = 3.61 eV, and I P of Cl = 17.422 eV
(A) 1.24 × 1023 atoms (B) 9.82 × 1020 atoms
(C) 2.02 × 1015 atoms (D) none of these
 The binding energy of an electron in the ground state of the He atom is equal to 24 eV. The energy required to remove both the electrons from the atom will be
(A) 59 eV (B) 81 eV
(C) 79 eV (D) None of these
 The wave number of the shortest wave length transition in Balmer series of atomic hydrogen will be
(A) 4215 Å (B) 1437 Å
(C) 3942 Å (D) 3647 Å
ANSWERS TO ASSIGNMENT PROBLEMS
Subjective:
Level – O
 (i) Total electrons in n = 4 are 2n2 i.e. 2 × 42 = 32. Half of them have ms = −1/2.
(ii) n = 3, means 3s orbitals which can have 2 electrons.
 Nitride ion is N3−.
P = 7, e = 10, n = 7
 Refer to theory.
 Mn4+ (Z = 25) = 1s22s2sp63s23p63d3. It has three unpaired electrons.
Cr3+ (Z = 24) = 2s22s22p63s23p63d3. It also has three electrons.
 (i) 1s (ii) 2px or 2py
 Refer to theory.
 We know that λ = h/mv, since mass of the car is large hence wavelength is very small.
 Anode rays are beams of positive ions which are different for different gases.
 In the presence of magnetic field the orbitals present in a subshell take up different orientations.
 Number of lines
 4.77 × 10−8
 2 × 10−3
 Mass = 5.486 × 107 kg and charge = 9.65 × 104 C
 (a)
(b)
 Cathode rays are same in both cases, but anode rays will be different.
 Refer to theory.
 Emission spectrum, Lyman series.
Level – I
 2.
 4. − 0.545×1011 erg
 1.09 × 108 cm/sec 6. 2.01 ×1018
 λ = 1215.68 A° 8. 364.4 nm
 Z = 3 10. V3 = 7.275 × 105 m/sec
rps = 2.432 × 1014
 (a) 3.313 × 10–33m
(b) 7.25 × 10–7m
(c) 1.227Å
 r1= 0.53 Å : r1= 0.265 Å
r2 = 2.12 Å for hydrogen : r2 = 1.06 Å for He+
r3 = 4.77 Å : r3 = 2.385 Å
 19.5 kcal/mole 14. 6630 nm
 5.77 × 10−4
Level – II
 (a) 3.67 × 10–5 cm
(b) 30.6 eV for Li2+, 54.4 eV for Be3+
 27.2 × 1018
 (a) 1.86 × 104 eV
(b) 4.85 × 10–11m
 (i) 113.7 Å
(ii) 3
 6603 Å
 (i) 628.72 × 1021 atoms
(ii) 832.50 kJ.
 λ1 = 1028 Å, λ2= 1216 Å, λ3 = 6558 Å
 5.79 × 105
 (i) 3, (ii) 108.87 eV, (iii) 1.01 × 10–6cm (iv) 122.4 eV
 (a) 2 and 3
(b) 5
(c) 12.09 eV and 1.9 eV
Objective:
Level –I
 A 2. C 3. A
 A 5. B 6. A
 C 8. C 9. A
 B 11. A 12. C
 B 14. B 15. A
 A, C 17. A, C 18. B, C
 A, B, C 20. C, D 21. D
 B 23. B 24. B
 C
Level – II
 A 2. D 3. A
 C 5. C 6. D
 C 8. A 9. A
 B 11. B 12. B
 A 14. A 15. B
 A 17. D 18. A
 C 20. D