ATOMIC STRUCTURE_1

 IIT–JEE Syllabus
Atomic structure; Rutherford Model; Spectrum of hydrogen atom; Bohr model;
de Broglie relations, Uncertainty principle, Quantum model; Electronic configuration of elements ( upto to atomic number 36); Aufbau principle, Pauli’s exclusion principle and Hund’s rule, shapes of s,p, and d orbitals.
 Dalton’s Atomic Theory
All the objects around you, this book, your pen or pencil and things of nature such as rocks, water and plant constitute the matter of the universe. Matter is any substance which occupies space and has mass.
Dalton, in 1808, proposed that matter was made up of extremely small, indivisible particles called atoms. (In Greek atom means which cannot be cut). This concept was accepted for number of years.
The main postulates of Dalton’s atomic theory are
 Matter is made up of small indivisible particles, called atoms.
 Atoms can neither be created nor destroyed. This means that a chemical reaction is just a simple rearrangement of atoms and the same number of atoms must be present before and after the reaction.
 Atom is the smallest particle of an element which takes part in a chemical reaction.
 Atoms of the same element are identical in all respects especially, size, shape and mass.
 Atoms of different elements have different mass, shape and size.
 Atoms of different elements combine in a fixed ratio of small whole numbers to form compound atoms, called molecules.
However, the researches done by various eminent scientists and the discovery of radioactivity have established beyond doubt, that atom was not the smallest indivisible particle but had a complex structure of its own and was made up of still smaller particles like electrons, protons, neutrons etc. At present about 35 different subatomic particles are known but the three particles namely electron, proton and neutron are regarded as the fundamental particles.
We shall now take up the brief study of these fundamental particles. 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 and high voltage, which produces cathode rays consisting of negatively charged particles, named as electrons. The e/m ratio for cathode rays is fixed whose value is 1.76 ´ 10^{8} C/g
We know that an atom is electrically neutral, if it contains negatively charged electrons it must also contain some positively charged particles. This was confirmed by Goldstein in his discharge tube experiment with perforated cathode. On passing high voltage between the electrodes of a discharge tube it was found that some rays were coming from the side of the anode which passed through the holes in the cathode. These anode rays (canal rays) consisted of positively charged particles formed by ionization of gas molecules by the cathode rays. The charge to mass ratio ( e/m value) of positively charge particles was found to be maximum when the discharge tube was filled with hydrogen gas as hydrogen is the lightest element. These positively charged particles are called protons.
e/m varies with the nature of gas taken in the discharge tube. The positive particles are positive residues of the gas left when the gas is ionized.
The neutral charge particle, neutron was discovered by James Chadwick by bombarding boron or beryllium with a–particles.
Characteristics of the three fundamental particles are:
Electron  Proton  Neutron  
Symbol  e or e^{–}  p  n 
Approximate relative mass  1/1836  1  1 
Approximate relative charge  –1  +1  No charge 
Mass in kg  9.109´10^{–31}  1.673´10^{–27}  1.675´10^{–27} 
Mass in amu  5.485´10^{–4}  1.007  1.008 
Actual charge (coulomb)  1.602´10^{–19}  1.602´10^{–19}  0 
Actual charge (e.s.u.)  4.8 ´ 10^{–10}  4.8 ´ 10^{–10}  0 
The atomic mass unit (amu) is 1/12 of the mass of an individual atom of _{6}C^{12}, i.e., 1.660´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 while the neutron is not charged.
3. 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.
3.1. Thomson’s Model
J.J. Thomson, in 1904, proposed that an atom was a sphere of positive charge in which number of electrons are embedded. This model is called the plum – pudding model.
This model could not satisfactorily explain the results of scattering experiment carried out by Rutherford.
3.2. Rutherford’s Model
a– 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 a–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 a particle, as ZnS has the remarkable property of converting kinetic energy of a particle into visible light. [For this experiment, Rutherford specifically used a particles because they are relatively heavy resulting in high momentum].
Observation
 Majority of the a–particles pass straight through the gold strip with little or no deflection.
 Some a–particles are deflected from their path and diverge.
 Very few a–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”]
Conclusions
 The fact that most of the a – particles passed straight through the metal foil indicates the most part of the atom is empty.
 The fact that few a – particles are deflected at large angles indicates the presence of a heavy positively charge body i.e., for such large deflections to occur a – 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.
Atomic model
On the basis of the above observation, Rutherford proposed an atomic model as follows.
 All the protons (+ve charge) and the neutrons (neutral charge) i.e 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^{–13}cm while the atom has a diameter of the order of 10^{–8 } So, the size of atom is 10^{5} 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 in various circular orbits.
 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 of Rutherford’s atomic model
 Position of electrons: The exact positions of the electrons from the nucleus are not mentioned
 Stability of the atom: Neils Bohr pointed out that Rutherford’s atom should be highly unstable. According to the law of electro–dynamics, the electron should therefore, continuously emit radiation and lose energy. As a result of this a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus.
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.
Atomic terms
(a) Atomic Number (Z)
The atomic number of an element is the number of protons contained in the nucleus of the atom of that element.
(b) Nucleons
Protons and neutrons are present in a nucleus. So, these fundamental particles are collectively known as nucleons.
(c) Mass Number (A)
The total number of protons and neutrons i.e, the number of nucleons present in the nucleus is called the mass number of the element.
(d) Nuclides
Various species of atoms in general. A nuclide has specific value of atomic number and mass number.
IUPAC notation of an atom (nuclide)
Let X be the symbol of the element. Its atomic number be Z and mass number be A. Then the element can be represented as
(e) Isotopes
Atoms of the element with same atomic number but different mass number e.g. _{1}H^{1}, _{1}H^{2}, _{1}H^{3}. There are three isotopes of hydrogen.
(f) Isobars
Atoms having the same mass number but different atomic numbers, e.g. _{15}P^{32} and _{16}S^{32} are called isobars.
(g) Isotones
Atoms having the same number of neutrons but different number of protons or mass number, e.g. _{6}C^{14}, _{8}O^{16}, _{7}N^{15} are called isotones.
(h) Isoelectronic
Atoms, molecules or ions having same number of electrons are isoelectronic e.g. N_{2},CO, CN^{–}.
(i) Nuclear isomers
Nuclear isomers (isomeric nuclei) are the atoms with the same atomic number and same mass number but with different radioactive properties.
Example of nuclear isomers is
Uranium –X (half life 1.4 min) and
Uranium –Z (half life 6.7 hours)
The reason for nuclear isomerism is the different energy states of the two isomeric nuclei.
Other examples are
_{ 30}Zn^{69} _{30}Zn^{69}
(T_{1/2} = 13.8 hr) (T_{1/2} = 57 min)
_{ 35}Br^{80} _{35}Br^{80}
(T_{1/2} = 4.4 hour) (T_{1/2} = 18 min)
(j) Isosters
Molecules having same number of atoms and also same number of electrons are called isosters.
E.g., (i) N_{2} and CO
(ii) CO_{2} and N_{2}O
(iii) HCl and F_{2}
(k) Atomic mass unit
Exactly equal to 1/12 of the mass of _{6}C^{12} atom
1 amu = 1.66 ´10^{–27} kg = 931.5 MeV
4. 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 (l), frequency (n), velocity (c), wave number and amplitude (a).
Ordinary light rays, X–rays,g–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.
(i) Wave length (l)
The distance between two neighbouring troughs or crests is known as wavelength. It is denoted by l and is expressed in cm, m, nanometers (1nm=10^{–9}m) or Angstrom (1Å=10^{–10}m).
(ii) Frequency (n)
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 n(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 (l)
n µ or n =
(iii) 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 = nl or l =
(iv) Wave number
It is defined as number of wavelengths per cm. It is denoted by and is expressed in cm^{–1}.
= (or) =
(v) 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.
Wavelengths of electromagnetic radiations
Electromagnetic radiations  Wave length (Å) 
Radio waves  3´10^{14} to 3 ´10^{7} 
Micro waves  3´10^{9} to 3 ´10^{6} 
Infrared (IR)  6´10^{6} 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 
5. 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.
Hydrogen Atom
If an electric discharge is passed through hydrogen gas taken in a discharge tube under low pressure, and the emitted radiation is analysed 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
l = wave length
R = Rydberg constant (109678 cm^{–1})
n_{1} and n_{2} have integral values as follows
Series n_{1} n_{2} 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
Barckett 4 5,6,7 etc Infra – red
Pfund 5 6,7,8, 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.
Types of Emission spectra
 i) Continuous spectra: When white light from any source such as sun or bulb is analysed 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 continous that each of them merges into the next. Hence the spectrum is called as continuous spectrum.
 ii) 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.
6. 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
 The energy of a quantum is directly proportional to the frequency of the radiation . E µ n (or) E = hn were n 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 hn, 2hn,3hn………..nhn. where n is the positive integer.
Neils Bohr used this theory to explain the structure of atom.
7. 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, (DE) is given by
DE = E_{2} – E_{1} = hn
where E_{2} and E_{1} are the energies of the electron in the first and second energy levels, and n 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 appear 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 Î_{o} is permittivity of free space)
K = 9 ´10^{9} Nm^{2} C^{–2}
In C.G.S. units, value of K = 1 dyne cm^{2} (esu)^{–2}
The centrifugal force acting on the electron is
Since the electrostatic force balance the centrifugal force, for the stable electron orbit.
= … (1)
(or) v^{2} = … (2)
According to Bohr’s postulate of angular momentum quantization, we have
mvr =
v =
v^{2} = … (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 r_{o}.
r_{o} = =
= 5.29 ´10^{–11} m = 0.529 Å
Radius of n^{th} orbit for an atom with atomic number Z is simply written as
r_{n} = 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 mv^{2}
Potential energy =
Total energy = 1/2 mv^{2 }– … (4)
From equation (1) we know that
=
\ ½ mv^{2} =
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^{–19}J)
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 ¾® H^{+}+ e^{–} (ionisation).
Exercise 1: Calculate the ratio of K.E and P.E of an electron in a orbit
Exercise 2: If the energy of third shell of hydrogen atom is –45 a.u. (arbitrary units) what would be its ionisation potential?
Explanation for hydrogen spectrum by Bohr’s theory
According to the Bohr’s theory electron neither emits nor absorbs energy as long as it stays in a particular orbit. However, when an atom is subjected to electric discharge or high temperature, and electron in the atom may jump from the normal energy level, i.e., ground state to some higher energy level i.e, exited state. Since the life time of the electron in excited state is short, it returns to the ground state in one or more jumps.
During each jump, energy is emitted in the form of a photon of light of definite wavelength or frequency. The frequency of the photon of light thus emitted depends upon the energy difference of the two energy levels concerned (n_{1}, n_{2}) and is given by
hn = E_{2} – E_{1}
=
n =
The frequencies of the spectral lines calculated with the help of above equation are found to be in good agreement with the experimental values. Thus, Bohr’s theory elegantly explains the line spectrum of hydrogen and hydrogenic species.
Bohr had calculated Rydberg constant from the above equation.
n = =
where = 1.097 ´ 10^{7}m^{–1} or 109678 cm^{–1}
i.e., Rydberg constant (R)
\ = RZ^{2}
Illustration 1: What is the wavelength of second line of Paschen series of Be^{3+} ion?
Solution: Z = 4 for Beryllium
For Paschen series n_{1} = 3
For 2^{nd} line n_{2} = 5
\
= 109678 × 4^{2}
= 109678 × 4^{2} [0.0711]
= 124769.6 cm^{–1}
l = 8.013 × 10^{–6} cm
Exercise –3: How many spectral lines are emitted by atomic hydrogen excited to
n^{th} energy level?
Exercise – 4: The electron in a hydrogen atom revolves in the third orbit. Calculate (i) the energy of the electron in this orbit (ii) the radius of the third orbit and (iii) the frequency and wavelength of the spectral line emitted when the electron jumps from the third orbit to the ground state.
(Given m_{e} = 9.1 ´ 10^{–28}g; e = 4.8 ´ 10^{–10} esu; h = 6.6 ´ 10^{–27} erg sec
Illustration – 2: A doubly ionised lithium atom is hydrogen like with an atomic no. 3.
 i) 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)?
 ii) How many spectral lines are observed in the emission spectrum of the above excited system?
Solution: i) Li^{2+} is a Hlike (1s^{1}) particle, whose energy levels are given by
E_{a} = [for Li, Z (atomic weight) = 3]
=
The energy required to excite the electron from the I to the III orbit is
E_{3} – E_{1} = = 108.8 ´ 1.6 ´ 10^{–19} J
Q 1 eV = 1.6 ´ 10^{–19} J
Therefore, l =
= 113.7 ´ 10^{–10} m = 113.7Å
 ii) Three spectral lines will be observes in the emission spectrum of the above excited system corresponding to the transition shown below.
3 ® 1 3 ® 2 2 ® 1
 of spectral lines = = 3
Exercise – 5: Calculate the ionisation potential of a hydrogen atom when R_{H} = 109600 cm^{–1}. Hence calculate energy required for the transition n = 4 to n = 2 of He^{+} ion.
Exercise6: How much energy will be emitted when electrons in a gram atom of hydrogen falls from third Bohr orbit to first?
Exercise7: A photon of wavelength 300nm causes the emission of two photons. If the wavelength of one of them is 700 nm. What is the wavelength of the other photon?
Calculation of velocity
We know that
mvr =
v =
By substituting for r we are getting
v =
Where excepting n and z all are constants
v = 2.18 ´10^{8} cm/sec.
Further application of Bohr’s work was made, to other one electron species (Hydrogenic ion) such as He^{+} and Li^{2+}. In each case of this kind, Bohr’s prediction of the spectrum was correct.
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
Exercise – 8: An electron in Bohr’s hydrogen atom has energy of – 0.85 eV. Calculate the angular momentum of the electron.
Illustration 3: Consider the hydrogen to be a proton embedded in a cavity of radius a_{0} (Bohr radius) whose charge is neutralized by the addition of an electron to the cavity in vacuum, infinitely slowly. Estimate the average total energy of an electron in its ground state in a hydrogen atom as the work done in the above neutralization process. Also if the magnitude of the average kinetic energy is half the magnitude of the average potential energy, Find the average potential energy?
Solution: Given that total energy is the work done in the neutralisation process i.e. moving electron infinitely, slowly into the cavity.
T.E. =
TE = PE + KE
Let PE = – x
KE =
TE = –x +
\ PE = x = 2 × TE
=
=
Exercise 9: Calculate l 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.
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.
(i) Principal quantum number (n)
It tells the main shell in which the electron resides and the approximate distance of the electron from the nucleus. It also tells the maximum number of electrons a shell can accommodate is 2n^{2}, 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
(ii) Azimuthal or angular momentum quantum number (l)
This represents the number of subshells present in the main shell. These subsidiary orbits within a shell will be denoted as 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’ .
(iii) 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 –l to +l including zero. Thus m can be –1, 0, +1 for l = 1
(iv) The spin quantum number (s)
Just like earth 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.
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 l, 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 (l =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 l = 1. Hence, m = –1, 0, +1. Thus, p–orbital have three different orientations. These are designated as p_{x},p_{y} & p_{z} depending upon whether the density of electron is maximum along the x y and z axis respectively. As they are 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 p orbitals belonging to a particular energy shell have equal energies and are called degenerate orbitals.
d–orbitals (l =2)
For d–orbitals, l =2. Hence m=–2,–1,0,+1,+2. Thus there are 5 d orbitals. They have relatively complex geometry. Out of the five orbitals, the three (d_{xy}, d_{yz},d_{zx}) project in between the axis and the other two and lie along the axis.
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 arrows 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+l) rule.
The energy of an orbital depends upon the sum of values of the principal quantum number (n) and the azimuthal quantum number (l). This is called (n+ l) rule.
According to this rule,
“In neutral isolated atom, the lower the value of (n+ l) for an orbital, lower is its energy. However, if the two different types of orbitals have the same value of
(n+ l), the orbitals with lower value of n has lower energy’’.
Illustration of (n + l) rule
Type of orbitals  Value of n  Values of l  Values of (n+ l)  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=3 
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 ¯
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 untill 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 minimised 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.
Illustration 4: We know the Hund’s rule. How do we 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.
2p_{x} 2p_{y} 2p_{z}
  
½ ½ ½ = 1½
The maximum multiplicity means that the total spin of unpaired electrons is maximum.
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.
(a) Each orbital can accommodate two electrons
(b) The number of electrons to be accomodated in a subshell is 2 ´ number of
degenerate orbitals.
Subshell  Maximum number of electrons 
s  2 
p  6 
d  10 
f  14 
(c) The maximum number of electron in each shell (K,L,M,N…) is given by 2n^{2}.
where n is the principal quantum number.
(d) The maximum number of orbitals in a shell is given by n^{2} where n is the principal quantum number.
(e) The number of nodal planes associated with an orbital is given by l values.
Importance of knowing the electronic configuration.
The chemical properties of an element are dependent on the relative arrangement
of its electrons.
Illustration 5: Write the electronic configuration of nitrogen (atomic number= 7).
Solution:
1s^{2}  2s^{2}  2p^{3}  
¯  ¯    
Exceptional Configurations
Stability of half filled and completely filled orbitals
Cu has 29 electrons. Its expected electronic configuration is
1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{9}
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
1s^{2} 2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}
Fully filled and half filled orbitals are more stable.
Exercise 10: We know that fully filled and half filled orbitals are more stable. Can you write the electronic configuration of Cr(Z = 24)?.
Exercise 11: How many 7s electrons are there in an atom with Z = 104?
Atomic Structure
Hints to Subjective Problems
Level – I
 Longest wavelength is associated with least energy.
 Total energy emitted = number of photons emitted × energy of 1 photon
 At the closest distance of approach, the alpha particle will have the same kinetic and potential energies
 Period of revolution = T =
Level – II
 a) KE for a H – like atom is same as the magnitude of total energy
 b) l µ 1/m if the velocity is same.
 Use quantization condition for a Bohr orbit in order to evaluate the n values.
 The energy of the incident light should always be greater than the work function for photo electric emission to take place.
Level – III
 Identify the wavelengths corresponding to the transitions; which one is shorter or larger. Using the appropriate evaluate the work function and hence stopping potential.
 Evaluate the uncertainty in position with the help of radius of first orbit of He^{+} ion
 d) If the energy of the incoming electron is less than DE then it cannot interact with the atom and if the energy is more then transition is made from one level to another and the remaining energy will be in the form of K.E.
 One Rydberg = IP of H atom.
 Energy of incident photon will be partially utilized in breaking the I_{2} molecule and residual energy in the form of K.E.
 Removal of electron involves absorption of energy and addition of electron involves release of energy.
Energy absorbed by F + Energy absorbed by Cl = Total energy absorbed
Energy released by F + Energy released by Cl = Total energy released
 Using quantization condition and the condition for the stability of the electron in the circular orbit expression for r can be derived.
Solutions to Subjective Problems
Level – I
For He; Z = 2; For Paschen series n_{1} = 3
For longest wavelength n_{2} = 4
= 109678×4×
= 109678 ×4×
l = 4689 A^{o}
 wave number of first line of Balmer, =
\wave length of first line of Balmer =
wave number of ultimate line of Balmer, = =
\wave length of ultimate line of Balmer =
\Ratio =
 Energy corresponding to 60 W bulb is
E = 60 J/sec = 60 ×3600 J/hr
For 10 hours the energy is
E = 60×3600×10
Energy corresponding to the photon =
= 3.37×10^{19} J
\ No. of photons emitted =
= 6.404 ×10^{24}
 From equation E_{n} = , the energy change of a H–atom that goes from initial state of quantum number n_{i} to a final state of quantum number n_{f} is
DE = E_{f} – E_{i} =
Hence n_{i} = 1, n_{f} = 3, E_{1} = –13.6 eV
DE = –13.6 = 12.08 eV.
 (a) From equation r_{n} = n^{2}a_{o}
n = » 435
(b) E_{n} =
(Such an atom would obviously be extremely fragile and can be easily ionized).
 At the distance of closest approach of an alpha particle (r)
P.E. = K.E.
v^{2} =
=
v= 6.35 ×10^{6} m/sec
= =
1.54 =
V = 63.24 volts
 8. Dx × Dp ³
(Dp)^{2} = (Dx)^{2} = (since Dp = Dx)
Dp = =
mDv = Dp =
Dv =
m = 9.1 ×10^{31} Kg
\ Dv = 7.98 × 10^{12} m/sec
 n= = = 5.45 ´10^{17} s^{–1}
½ mv^{2} = hn–hn_{o}
= (6.63´10^{–27} ergs–sec) (5.45´10^{17}s^{–1}) –2.62 ´10^{–9 }ergs
K.E. = 9.93 ´10^{–10} ergs
 Let the number of photons required = n
n
n = = = 27.6 = 28 photons
 Energy corresponding to the incident photon
E = 6.626 × 10^{34} × 3.2 × 10^{16}
= 2.12 ×10^{17} J
K.E of photoelectron = th energy of photon
= ×2.12×10^{17}
= 1.590 × 10^{17} J
hn = hn_{0} + KE
2.12 × 10^{17} = hv_{0} + 1.590 ×10^{17}
hv_{0} = 5.297 × 10^{18 }J
n_{0} =
= 7.995 10^{15} Hz
 Time taken for one complete revolution is given by T =
T a n^{3}
, n^{2} = 4 Þ n = 2
\ the ratio of time taken = 1^{3} : 2^{3}
= 1:8
 For nth orbit mvr =
= mv =
Again from deBeoglie’s equation we have
\
l =
 Energy of a single photon = h n =
= = 4.42 ´10^{–19}J …(1)
Energy emitted by the bulb = ´150 J = 12 J …(2)
Let ‘n’ photons are emitted per second
n´4.42´10^{–19}J = 12 J
n = 27.2 ´10^{18}
 According to Rydberg equation
= RZ^{2}
for lowest frequency in Lyman series n_{2} = 2 and n_{1} = 1
\ = 1.09678 ´10^{7} ´ (Z = 1 for hydrogen)=
l= 1.2156´10^{–7} metre
l = 1215.6 Å
Frequency = n = =
= 2.4679´10^{15}H_{z}
Energy E = hn = 6.626´10^{–34} ´2.4679´10^{15}
= 16.352´10^{–19} J = eV = 10.2 eV
Now energy for corresponding spectral line of Li^{2+} species
= Z^{2}E_{H} (for Li^{2+}, Z =3)
= (3)^{2} ´10.2 = 91.80 eV
Level – II
 Energy corresponding to the photon =
= 3.822 ×10^{19} J
the power output = 2.79 Watt/m^{2}
= 2.79 J/sec/m^{2}
\no. of photons emitted per sq. meter sec =
= 7.30 ×10^{18}
 i) KE of an electron in the orbits of hydrogen and hydrogen like atoms
=
= 3.4 eV
 ii) =
= 6.654 × 10^{10} m
= 6.654 A^{o}
 a) Energy of photon , E = W + KE
= 1.82 + 0.73 = 2.55 eV
 b) The energy of an electron in n^{th} orbit of hydrogen atom is given as
E_{n} =
E_{1} = –13.6 eV
E_{2} =
E_{3} = –1.51eV
E_{4} = –0. 85 eV
Obviously, the transition 2 ¬ 4 leads to the release of photon of energy
2.55 eV as
QDE =  E_{2} – E_{4} = 2.55 eV
 If l_{o} be the threshold wavelength and f be the work function
l_{o} = = .
For copper = l_{o} = = 276 nm
For sodium = l_{o} = =540 nm
For caesium = l_{o} = = 654 nm
 a) =
= 1.46 × 10^{9} m
 b)
when velocity is same l µ
 a) Energy corresponding to 8205.8 A^{o} =
= 2.422 × 10^{19} J
= 1.572 eV
1.512 eV = E_{1H} × (1)^{2} ×
1.512 eV =
E_{1 H} = 13.608 eV
\ Ionisation energy of hydrogen atom = 13.6 eV
 b)
= 916 A^{o}
 Angular momentum of the electron = 5.2728 × 10^{34} kgm^{2}/sec
= 5.272 × 10^{34}
n_{1} = 4.99 » 5
Angular momentum of electron after transition = 3.1636 × 10^{34} Kgm^{2}/sec
= 3.1636 × 10^{34}
n_{2} = 2.999 » 3
\ the electron makes trasition from . 5 ® 3
n = = c × R_{h} × Z^{2}
n = 3 × 10^{10} × 109678 × 3^{2} ×
= 2.1058 × 10^{15} Hz
 Energy corresponding to n = 2 =
= 5.425 × 10^{12} erg
5.425 × 10^{12}
l = 3.66 × 10^{5} cms
n_{2} = 4
next level is n = 5
velocity of electron in 5^{th} level of He^{+} =
= = 87.39 ×10^{6} cm/sec
velocity of the electron in 5^{th} level : velocity of light = 1:344
 Lowest energy in u.v. region is
E = hc R_{H}
_{ }6.626 ×10^{34} × 3 ×10^{8} × 1.09 × 10^{7} ×
= 1.625 × 10^{18} J
This is energy corresponding to 1 atom of H
For 1 g atom the energy is
E = 1.625 × 10^{18} × 6.023×10^{23} J
= 978.753 KJ mole^{1}
 eV_{o} = –f f = work function
f = = 3.63 eV
K.E. = hn – f
= = 2.57 eV
\ Stopping potential = 2.57 Volts
 a) E = hn
n = = 5.99 ´10^{14} sec^{–1}
 b) Energy associated with photon of blue light with wave length 450 nm or 450 ´10^{–9} m
E = =
= 4.417 ´10^{–19} J/ Atom
Above energy is greater than threshold energy (i.e, 3.97 ´10^{–19}J).
Hence electron will be ejected.
 Kinetic energy of electron = hn – hn_{critical}
= –
KE = = 1.22 eV
(Since 1240 nm = 1eV)
 14. Energy required to stop the ejection of electron is equal to kinetic energy i.e., (½ mv^{2})
Given K.E = 0.24 eV
l of incident photon = 253. 7 nm
hn = hn_{o} + K.E
Work function (hn_{o}) = hn – K.E = – ½ mv^{2}
= = 4.89 – 0.24
= 4.65 eV
 Let the number of moles of I_{2} dissociated are n
Total energy = 1 kJ
= 1000 Joule
N_{A} ® Avogadro’s no.
On solving above equation we get
n = 4.17 ´10^{–3} moles
level – iiI
 a) Q 1eV = 1.602 ´ 10^{–12} erg
Also DE = = E_{3} – E_{2} = R_{H}.c.h.Z^{2}
or 16.52 ´ 1.602 ´ 10^{–12} = 109678 × Z^{2} ×´ 3 ´ 10^{10} ´ 6.626 ´ 10^{–27}
\ Z^{2} = 8.74 \ Z » 3
 b) DE = E_{4} – E_{3} = R_{H}.c.h.Z^{2}
= 109678 ´ 3 ´ 10^{10} ´ 6.626 ´ 10^{–27} ´ 3^{2} ´ = 1.744 ´10^{–10} erg
= 108.87 eV
 c) = 109678 ´ 9
l = 1.01 ´ 10^{–6} cm
 d) E. = mu^{2 } = m
= = 1.962 ´ 10^{–10} erg
= 122.4 eV
 Let wavelength for the transition 5®4 and be that of 4 ® 3 transition
= = 4.502 × 10^{7} m
= 450.2 mm
= 2.089 × 10^{7} m = 208.9 nm
shorter wavelength is , longer wavelength is
stopping potential for the photo electrons ejected by the light of wavelength
l_{2} is 3.95 Volts
Energy corresponding to l_{2 }is
= 5.95 eV
since stopping potential = 3.95 V;
K.E. = 3.95 eV
E = f + KE_{1}
5.95 = f + 3.95
f = 2.00 eV
Energy corresponding to l_{1} is
= 2.75 eV
E_{1} = f + K.E_{2}
2.75 = 2.00 + KE_{2}
KE_{2} = 0.75 eV
\stopping potential = 0.75 V
 Total energy liberated during transition of electron from nth shell to first excited state (i.e., 2^{nd} shell) = 10.20 + 17.0 = 27.20 eV = 27.20 ´ 1.602 ´ 10^{–12} erg
Q
\ 27.20 ´ 1.602 ´ 10^{–12} = R_{H} ´ Z^{2} ´ h ´ c … (1)
Similarly, total energy liberated during transition of electron from nth shell to second excited state (i.e., 3^{rd} shell) = 4.25 + 5.95 = 10.20 eV
= 10.20 ´ 1.602 ´ 10^{–12} erg
\ 10.20 ´ 1.602 ´ 10^{–12} = R_{H} ´ Z^{2} ´ h ´ c … (2)
Dividing Eq. (1) by Eq. (2)
n = 6
On substituting the value of n in Eqs. (1) or (2).
Z= 3
 Radius of 1^{st} orbit of the He^{+} ion =
= 0.265 = 0.265 × 10^{10} m
Uncertainty in the location = 1 % of the radius of the 1^{st} orbit
= 0.265 × 10^{12} m
Dp × Dx ³
Dv ³
³
³ 2.18 × 10^{8} m/sec
velocity of the electron
v = m/sec
= 4.368 ×10^{6} m/sec (since Z = 2, n = 1)
 a) I.P. of the atom is
I.P. = E_{¥} – E_{1}
= 0 – (15.6) = 15.6 eV
 b) Series limit at n = 2 is
E = E_{¥} – E_{2}
= 0(5.3) = 5.3 eV
= 5.3 eV
= 5.3 eV
l = 233.9 nm
 c) D E for the transition n = 3 ® n = 1 is
D E = E_{3} – E_{1}
= 3.08 – (15.6)
= 12.52 eV
= 2.005 ×10^{18} J
= 2.005 ×10^{18}
= 1.009 × 10^{7} m^{1}
 d) The difference in energy between n = 1 and n = 2 is 10.3 eV
 i) Since the incoming electron has 6 eV energy, it will not be able to do anything hence it will not interact with the atom.
 ii) In case it has 11 eV energy, it will impact 10.3 eV energy to this atom and (11 eV – 10.3 eV) 0.7 eV energy is carried by this electron. Due to transfer of 10.3 eV of energy it will jump from n = 1 to n = 2 level.
 1 g H contains = N atoms
\ 1.8 g contains = N ´ 1.8 atoms
= 6.023 ´ 10^{23} ´ 1.8
= 10.84 ´ 10^{23} atoms
(a) \ No. of atoms in III shell =
= 292.68 ´ 10^{21} atoms
\ No. of atoms in II shell =
= 162.6 ´ 10^{21} atoms
and No. of atoms in I shell =
= 628.72 ´ 10^{21} atoms
(b) When all the atoms return to I shell, then
E¢ = (E_{3} – E_{1}) ´ 292.68 ´ 10^{21}
= ´ 1.602 ´ 10^{–19} ´ 292.68 ´ 10^{21}
= 5.668 ´ 10^{5} joule
E¢¢ = (E_{2} – E_{1}) ´ 162.6 ´ 10^{21}
= ´ 1.602 ´ 10^{–19} ´ 162.6 ´ 10^{21}
= 2.657 ´ 10^{5} joule
E = E¢ + E¢¢ = 5.668 ´ 10^{5} + 2.657 ´ 10^{5} joule
= 832.50 kJ
 Energy of photon = kinetic energy of the photo electron and threshold preparation.
hn_{1} = KE_{1} + hn_{0} …(1)
hn_{2} = KE_{2} + hn_{0} …(2)
Multiplying (1) by 2 and substracting equation (2) from it
2hn_{1} – hn_{2} = hn_{0} (Q2KE_{1} = KE_{2})
2n_{1} – n_{2} = n_{0}
n_{0} = (1Å = 10^{–10} m)
n_{0} = 1.1483 ´ 10^{15} sec^{–1}
Also l = = 2.6126 ´ 10^{–7} m
= 2612.6Å
 Energy of I orbit of H like atom = 4R_{h}
= 4 ´ 2.18 ´ 10^{–18} joule
E_{1} for H = –2.18 ´ 10^{–18} J
Q E_{H like atom} = E_{1H} ´ Z^{2}
\ –4 ´ 2.18 ´ 10^{–18} = –2.18 ´ 10^{–18} ´ Z^{2}
\ Z = 2
i.e., Atomic no. of H like atom is 2 or it is He^{+} ion.
 a) For deexcitation of electron in He^{+} from n_{2} = 2 to n_{1} = 1
E_{2} – E_{1} =
Now E_{1} = 4R_{h}
\ E_{2} = – = –R_{h}
\ E_{2} – E_{1} = 3R_{h} = 3 ´ 2.18 ´ 10^{–18} J
\ E_{2} – E_{1} =
\ l = = 303.89 ´ 10^{–10 }m
= 303.89 Å
 b) Radius (r_{1}) of H like atom = = 2.645 ´ 10^{–9} cm
 Energy given to I_{2} molecule
^{ }= = = 4.417 ´ 10^{–19} J
Also energy used for breaking up to I_{2} molecule = = 3.984´10^{–19} J
\ Energy used in imparting kinetic energy to two I atoms
= [4.417 – 3.984] ´ 10^{–19} J
\ K.E./ iodine atoms = [(4.417 – 3.984)/2] ´ 10^{–19}
= 0.216 ´ 10^{–19} J
 Wave length emitted is in UV region and thus n_{1} = 1; For H atom
\
= 1.097 ´ 10^{7}
\ n = 2
Also the energy released is due to collision and all the kinetic energy is released in form of photon. Thus
=
or ´ 1.67 ´ 10^{–27} ´ u^{2} =
\ u = 4.43 ´ 10^{4} m sec^{–1}
 Let x and y be the amounts of F and Cl atoms in the given mixture. Since the ionisation of atoms absorbs energy, we can write.
x(6.023 ´ 10^{23} mol^{–1}) (27.91 ´ 10^{–22} kJ) + y (6.023 ´ 10^{23} mol^{–1}) (20.77 ´ 10^{–22} kJ)
= 272.2 kJ …(1)
Since the addition of electron to atoms releases energy, we can write
x(6.023 ´ 10^{23} mol^{–1}) (5.53 ´ 10^{–22} kJ) + y(6.023 ´ 10^{23} mol^{–1}) (5.78 ´ 10^{–22} kJ)
= 68.4 kJ …(2)
Solving for x and y, we get
x = 0.054 mol and y = 0.144 mol.
Percentage of F atoms = ´ 100 = 27.27
Percentage of Cl atoms = 100 – 27.27 = 72.73
 The basic expressions in Bohr model of the atom are as follows.
(a) Stability of the circular motion of the electron, i.e.
Attractive force = Centrifugal force
…(1)
Quantization of angular momentum
mvr = …(2)
Eliminating v in the above two expressions, we get
=
This gives r= n^{2} …(3)
 b) Now for the given problem
m = (200)m_{e} and Z = 3
Hence,
r = n^{2}
= n^{2}(8.856 ´ 10^{–14} m)
Equation Eq. (3) with the first Bohr orbit for the hydrogen atom, we get
n^{2}
or n^{2} = 200 ´ 3 or n = = 24
(c) The energy of the electron in the Bohr model of atom is
E = KE + PE = mn^{2} –
Using Eq. (I) , we get
E =
Substituting the expression of r from Eq. (3), we get
E =
=
= (3.899 ´ 10^{–15} J)
Hence, for the transition 1 ¬ 3, we get
DE = (3.899 ´ 10^{–15}J) = 3.465 ´ 10^{–15} J
l =
= 5.736 ´ 10^{–11} m = 57.36 pm
 O_{2} ¾® O_{N} +
O_{2} ¾® O_{N} + O_{N}
E = 498 ´10^{3} J / mole
= per molecule = 8.268 ´10^{–19} J
Energy required for excitation = 1.967 eV
= 3.146 ´10^{–19}J
Total energy required for photochemical dissociation of O_{2}
= 8.268 ´10^{–19} + 3.146 ´10^{–19 } = 11.414 ´ 10^{–19} J
= 11.414 ´10^{–19} J
l =
= 1.7415 ´10^{–7} m = 1741.5 Å
 Energy absorbed =
=
= 5.52 ´ 10^{–11} erg
= 5.52 ´ 10^{–18} joule
Now this energy is used in overcoming forces of attraction between surface of metal and imparting velocity to electron, therefore,
E_{absorbed} = E used in attractive forces + Kinetic energy of electron
\ Kinetic energy = 5.52 ´ 10^{–18} – 7.52 ´ 10^{–19} joule = 47.68 ´ 10^{–19} joule
 Energy of light absorbed by one photon =
Let n_{1} photons are absorbed, therefore,
Total energy absorbed =
Now E of light reemitted out by one photon =
Let n_{2} photons are reemitted then,
Total energy reemitted out = n_{2} ´
As given E_{absorbed} ´ = E_{reemitted out}
´ n_{1} ´_{ = }n_{2 }´
_{ }\ = ´ =
\ = 1.25
Solutions to Objective Problems
Level – I
 Since average mass is M + 0.5
= M + 0.5
\ (B)
 For the last electron of Ga the configuration is 4p^{1}. The added electron enters the vacant p orbital making configuration 4p^{2}
For 4p electron n = 4 and l = 1
\ (A)
 We know that r_{n} = 0.529 Å, = 0.529 (3^{2}–2^{2}) Å
(For hydrogen Z = 1, n = 1)
= 0 .529 ´5´10^{–8}cm
= 2.645 ´10^{–8 }cm
\ (A)
 Total spin (s) of an atom is given by
s = n (Where n is number of unpaired electrons) = ´ 4 = 2
\(C)
 Orbital angular momentum =
for d, l = 2
orbital angular momentum =
=
\ (A)
 Hint: Isosters have same number of electrons and atoms
 Hint: 7+3×8+1 = 32
\ (B)
 12. For one molecule
E = hn
For 1.5 moles
E = 1.5 ´ 6.023 ´10^{23} ´hn
= 1.5 ´ 6.023 ´10^{23} ´6.626 ´10^{–34} ´ 7.5 ´10^{14 } = 4.48 ´10^{5} J
\ (C)
 Hint: n_{1} = 2 and n_{2} = 3, 4, 5, …… corresponds to Balmer series
 Wave number of a spectral line
Energy =
= 6.626 ´10^{–34} ´3´10^{8}´5´10^{5}
= 9.93 ´10^{–23}kJ
\ (B)
 = R_{H}
= 109678
= 109678 ´
l = = 1215.6 Å
\ (A)
 Hint: r µ
 Hint: E_{n} of H like species =
Level – II
 Hint: For lowest wave length, the energy difference between two energy states should be highest.
 Energy difference between 3^{rd} and 4^{th} I.P.’s is very high.
 Number of quanta absorbed =
=
= = 1.065 ´10^{22}
\ (A)
 = R_{H}
R_{H} =
l =
= 486.0 nm
\(C)
 One of the possible excited state of He^{+} ion should have energy
E_{n} = eV
\ E_{n} = = 6.04 eV
 Energy difference between 3^{rd} and 4^{th} I.P.’s is very high.
 The ion has configuration [Ar] 4s^{0}3d^{6}
Before losing the electrons it would have been [Ar] 4s^{2}3d^{7}
\ the ion is Co^{3+}
\ (C)
 E_{n} of H like species =
D E for H like species = D E for H × Z^{2}
For Be^{3+} Z = 4
D E = 10.2 ×16 = 163.2
\ (D)
 Hint: Isoelectronic species have same number of electrons.
 Hint: The order of energies of orbitals is
2s < 2p < 3p < 4s < 3d
The transition from higher to lower levels result in emission of light.
 … (i)
…(ii)
Dividing equation (I) by equation (ii), we get
, , \ Z = 2
\ He^{+} \ (A)
 l = h /mv
= = 1.060 ´ 10^{–36} m
\ (B)
 = R
= (1.09678 ´ 10^{7}) ´ 2^{2} ´ , = 4.3871 ´ 10^{7}
or l = 2.279 ´ 10^{–8} m
\(A)
 l = =
10^{–10} =
On solving we get
V = 0.0826 Volt
\(C)
 hc=
= 1.6203 ×10^{6} m^{1}
\ (B)