ATOMIC STRUCTURE_2
- 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 character-as wave and
as particle.
Derivation of de Broglie 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 = hn —————— (i) (according to the Planck’s quantum theory)
where n 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)
hn = mc2
But n = c/l
h = mc2
(or) l = 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.
l = h/mv (or) l=
where mv = p is the momentum of the particle.
9. 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 (l)
Therefore 2pr = nl where ‘n’ is an integer and ‘r’ is the radius of the orbit But l = h/mv \ 2pr = nh /mv (or) mvr = n h/2p |
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 2pr, (i.e., the circumference of the circle).
Thus, the number of revolutions per second is =
Exercise-12. A moving electron has 4.55 ´10-25 joules of kinetic energy. Calculate its wavelength (mass = 9.1 ´10–31 kg and h = 6.6´10–34 kg m2sec–1).
Illustration 6: 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.
Solution: According to de Broglie equation
lA = and lB =
=
But pB = ½ pA (given)
= = ½
lB = 2lA = 2 ´ 5´10–8 m = 10–7 m
Exercise 13: Calculate the de Broglie wavelength of a ball of mass 0.1kg moving with a speed of 60ms–1.
Illustration 7: To what effective potential a proton beam be subjected to give its protons a wavelength of 25.2 nm?
Solution: Energy in Joules = charge on the electron in coulomb × potential difference in volts.
E = eV
KE = eV
(mv)2 = 2meV; mv =
From Debroghie equation we have l =
\ l =
\ we have
25.2 × 10-9 =
V =1.299´10-6Volts
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 = hn | Energy = mv2 |
| 2. Wavelength = | Wavelength = |
[Note: We should never interchange any of the above]
10. 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 (Dx) and the uncertainty in the momentum (Dp = m.Dv where m is the mass of the particle and Dv is the uncertainty in velocity) is equal to or greater than h/4p where h is the Planck’s constant.
Thus, the mathematical expression for the Heisenberg’s uncertainty principle is simply written as
Dx . Dp ³ h/4p
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 ±l. 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 8: 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.
Dx. Dp =
Dx ´ (m.Dv) = h/4p
Dv = = ´
Dv = 5.80 ´ 1010 ms–1
This value is much higher than the velocity of light and
hence not possible.
Exercise 14: Calculate the product of uncertainties of displacement and velocity of a moving proton (mass = 1.66 × 10–27 kg).
11. 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 Ervin Schrodinger in 1926. This model describes the electron as a three dimensioinal 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 zy (psi) = wave function of the electron.
Significance of y The eave 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 y is negative. But the probability must be always positive and cannot be negative, it is thus, proper to use y2 in favour of y.
Significance of y2: y2 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.
12. Photo Electric 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 ionisation energy like Caesium, 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 Vo, 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 (hn). Suppose, the threshold frequency of light required to eject electrons from a metal is no, when a photon of light of this frequency strikes a metal it imparts its entire energy (hno) 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 no there is no ejection of electron. If the frequency of light is higher than no (let it be n), the photon of this light having higher energy (hn), will impart some energy to the electron that is needed to remove it away from the atom. The excess energy would give a certain velocity (i.e, kinetic energy) to the electron.
hn = hno + K.E
hn = hno + ½ mv2
½ mv2 = hn–hno
where, n = frequency of the incident light
no = threshold frequency
hno is the threshold energy (or) the work function denoted by f = hno (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 hno.
Illustration 9: What in the minimum error in position of an electron moving with a speed of 500 ms–1 measured to an accuracy of 0.006 % (mass of the electron = 9.1 × 10–31 kg) ?
Solution: Dp = mDv = 9.1 × 10–31 × 500 ×
= 2.73 × 10–32 kg ms–1
Dx =
= 1.932 × 10–3 m
- Solution to Exercises
Exercise-1: K.E. =
P.E. =
\ P.E. = –2K.E
\
Exercise-2: E3 = = – 45
E1 = (IP of hydrogen) = 45 × 32
= 405 a.u.
Exercise-3:
Thus the number of lines emitted from nth energy level
= 1+2+3+ …………. n–1 = S (n–1)
Sn =
\ S(n–1)= =
Number of spectral lines that appear in hydrogen spectrum when an electron de excites from nth energy level =
Exercise-4: i) Energy of the electron
En =
= –
= – 2.43 ´ 10–12 erg
- ii) Radius of the III orbit
rn =
= 4.75 ´ 10–8 cm
= 4.75 Å
iii) Frequency and wavelength
E3 = and
DE = E3 – E2 =
=
= 19.44 ´ 10–12 erg
iii) Frequency of radiation
DE = hn
Therefore v (frequency of radiation emitted)
= =
= 2.95 ´ 1015 Hz
Wavelength of radiation
C = nl
l = = 1.017 ´ 10–5 = 1017 ´ 10–8 cm
= 1017Å
Exercise- 5:
When electron is removed from 1st orbit of hydrogen atom n1 =1;
n2 = ¥
cm–1 =109600 cm-1
E =
= 6.626 × 10–27 × 3 ´ 1010 × 109600
= 2.17 × 10-11 erg
Energy required 4 ® 2 transition of He+ is
DE = hc × RH × Z2
DE = 2.17 × 10–11 × 22
DE = 2.17 × 10–11 ×
DE = 2.17 × 10–11 ×
DE = 1.63 × 10–11 erg
Exercise-6: E = (per atom)
= NA (per g atom)
= 1.09 × 107 m-1
= 9.68 × 106 m–1
E = 6.023 × 1023 × 6.626 × 10–34 × 3 × 108 × 9.68 × 106
= 1.160 × 106 J
Exercise-7:
l1 = 525 nm
Exercise-8: En =
E1 = 13.6eV
Þ – 0.85 =
n2 = 16
n = 4
for n = 4
angular momentum (mvr) =
= (n = 4) =
Exercise-9: E3 for H = –2.41 ´10–12 erg
E2 for H = –5.42 ´10–12 erg
For a jump from 3rd to 2nd shell
DE = E3–E2 =
l =
= = 6602.9 ´10–8 cm
= 6603 Å
Exercise-10: 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..
Exercise-11: The electronic configuration of the element with atomic number 104 is
1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p6,4d10,4f14,5s2,5p6,5d10,5f14,6s2, 6p6,6d2,7s2
Hence there are two 7s electrons.
Exercise-12: We are given,
Kinetic energy, mv2 = 4.55 ´10–25J
m = 9.1 ´10–31 kg
h = 6.6 ´10–34 kg m2s–1
= 6.6 ´10–34 J–sec
½ ´(9.1´10–31) v2 = 4.55 ´10–25
v2 = = 106
or v = 103 m sec–1
l = h/mv =
l = 7.25 ´10–7 m.
Exercise- 13: l =
= 1.1 ×10–34m
= This is too small for ordinary observation
Exercise-14: Dx × Dv =
=
= 3.17 × 10–8 m2/sec
- Solved Problems
14.1 Subjective
Problem 1: Estimate the difference in energy between the I and II Bohr orbit for a hydrogen atom. At what minimum atomic number, would a transition from n = 2 to n = 1 energy level result in the emission of X-rays with l = 3´ 10–8 m. Which hydrogen atom like species does this atomic number correspond to?
Solution: For a hydrogen atom, the expression of energy difference between two atomic levels is
DE = RHhc
Hence,
DE2 ¬ 1 = (1.0967 ´ 107 m–1) (6.626 ´ 10–34 Js) ´
= 1.635 ´ 10–18 J
For hydrogen like species, the expression is DE = Z2 RHhc
= Z2 (1.09677 ´ 107m–1)
= (1.09677 ´ 107) = 4.0
The species is He+
Problem 2: Calculate the ionization energy of (a) one Li2+ ion, and (b) one mole of Li2+ ions. Given Rydberg constant R = 1.0974 ´ 107 m–1.
Solution: The expression of ionisation energy is
DE = RHZ2hc
For Li2+ ion, Z = 3. Hence
DE = (1.0974 ´ 107 m–1) (9) (6.626 ´ 10–34 Js) (3 ´ 108 ms–1)
= 1.964 ´ 10–17 J
for one mole of ions, we have
DE¢ = NADE
= (6.023 ´ 1023 mol–1) (1.964 ´ 10–17 J)
= 1.1829 ´ 107 J mol–1 = 11829 kJ mol–1
Problem 3: What is the speed and de Broglie wavelength of an electron that has been accelerated by a potential difference of 220 V?
Solution: The kinetic energy of the electron under a potential difference of 220 V is given by the expression.
Hence v = = 8.80 ´ 106 ms–1
Now, using de Broglie relation, we get,
l = = 8.26 ´ 10–11 m
Problem 4: You are given 3.01 × 1023 atoms of A. If half of the atoms of A transfer one electron to the other half of A atoms; 204.5 kJ must ba added. If all A– ions are subsequently converted to A+ ions an additional 366.5 kJ must be added. Determine the ionisation potential and electron affinity of A in electron volts.
Solution: Ionisation Potential of 1.505 ×1023 atoms (a) – Electron affinity of 1.505 × 1023 atoms (b)
= = 1.276 × 1024 eV …(1)
The difference in IP and EA is taken as the energy is released when electrons are taken up by half of the atoms given
For converting 1.505 ×1023 A– ions to A+ ions; energy is needed to be spent against EA and energy has to be spent for IP.
A(g)– ¾® A(g) + e–
A(g) ¾® A(g)+ + e–
––––––––––––
A(g)– ¾® A(g)++ 2e–
\ I.P. of 1.505 × 1023 atoms (a) + EA of 1.505 × 1023 atoms (b)
= = 2.287 × 1024 eV …(2)
From (1) & (2) we have
a – b = 1.276 × 1027 eV
a + b = 2.287 × 1024 eV
Solving the above two equations we get
I.P. of 1.505 × 1023 atoms a = 1.781 × 1024 eV
EA of 1.505 × 1023 atoms b = 0.505 × 1024 eV
Hence I.P. of A = eV = 11.83 eV atom–1
EA of A = eV = 3.35 eV atom–1
Problem 5: If a stationary proton and a-particle are accelerated through same potential, then calculate the ratio of their de-Broglie’s wavelengths.
Solution:
(mv)2 = 2meV
p = mv =
for proton pp =
for a-particle pa =
l = (from de-Broglie’s equation)
Given Va = Vp =
ma = 4 units
mp = 1 unit
=
Problem 6: (a) What change in molar energy would be associated with an atomic transition giving rise to a radiation of 1 Hg?
(b) What is the relationship between electron volt and the wavelength in nm of the energetically equivalent proton?
Solution: a) If each of NA atoms give off 1 Hg photon, then
DE = NA (hn) = (6.023 ´ 1023 mol–1) (6.26 ´ 10–34) (1s–1)
= 3.990 ´ 10–10 J mol–1
- b) First, frequency equivalent to 1 eV is determined from the Planck’s equation and then the wavelength.
n = = (Q 1 eV = 1.602 ´ 10–19J)
= 2.42 ´ 1014 s–1
l = = 1.24 ´ 10–6m
Therefore, 1 eV = (1.24 ´ 10–6 m) 109nm / m
= 1240 nm
Problem 7: The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV is 0.83 V. Find the wavelength of incident photon and also identify energy levels in the hydrogen atom which will limit this wavelength?
Solution: KEmax of the photoelectron = 0.83 eV (stopping potential being 0.83V)
Energy of the incident photon = 1.7 + 0.83 = 2.53 eV
= 2.53 eV (since 1240 nm = 1 eV)
l = = 490.11 nm
This wavelength corresponds to visible region of hydrogen spectrum;
Hence n1 = 2, let n2 = n
\ = RH =
Þ
Solving the above equation we get
n = 4
\ The transition is from n = 2 ¾®4
Problem 8: It has been found that gaseous Iodine molecules dissociate into separate atoms after absorption of light at wavelengths less than 4995Å. If each quantum is absorbed by one molecule of I2, what is the minimum input in kcal/mole, needed to dissociate I2 by this photo chemical process.
Solution: E (per mole) = NAhn
E = Na
= 239.5 kJ/mol = 57.1 kcal/mole
Problem 9: Find (i) the total number of neutrons and (ii) total mass of neutrons in 1 mg of H2O18 ?
Solution: i) Since 20g of H2O18 contains 6.023 × 1023 molecules
\ 1g (1 × 10–6g) contains molecules
= 3.0115 × 1016 molecules
3.0115 × 1016 molecules of H2O18 contains 6.023 × 1016 hydrogen atoms and 3.0115 × 1016 atoms of O18
Each hydrogen atom contains zero neutrons
Each O18 atom contains 18 – 8 = 10 neutrons
Total number of neutrons per molecule of H2O18 = 10
Total number of neutrons for 3.0115 × 1016 molecules = 3.0115 × 1017
- ii) Mass of a neutron = 1.675 × 10–27 kg
Mass of 3.0115 × 1017 neutrons = (1.675 × 10–27 × 3.0115 × 1017)
= 5.044 × 10–10 kg
Problem 10. A gas of identical H like atom has some atoms in the lowest (ground) energy level A and some atoms I a particular upper (excited) energy level ‘B’ and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy level by absorbing monochromatic light of photon energy 2.7 eV. Subsequently, the atoms emit radiations of only six different photon energies. Some of the emitted photons have energy of 2.7 eV. Some have more and some less than 2.7 eV.
- i) Find the principal quantum number initially excited level B?
- ii) Find the transition energy for the gas atoms?
iii) Find the maximum and the minimum energies of the emitted photons?
Solution. The electrons being present in I shell and another shell n1. These are excited to higher level n2 by absorbing 2.7eV and on deexcitation emits six l and thus excited state n2 comes to be 4.
Now E1 = ; ;
Since deexcitation leads to different l having photon energy 2.7 eV and thus, absorption of 2.7 eV energy causing excitation to 4th shell and then re-emitting photons of 2.7 eV is possible only when n1 = 2 (the deexcitation from 4th shell occurs in I,II and III shell.)
E4 – E2 = 2.7 eV
E4 – E3 < 2.7 eV
E4– E1 > 2.7 eV
\ since n1 = 2 (as obtained by discussion)
Also E4 – E2 = 2.7 eV
\ – = 2.7 eV
\ E1 = –14.4 eV
I.P. = 14.4eV
Emax = E4 – E1 =
= + 14.4 = 13.5 eV
Emin = E4 – E3
= = 0.7 eV
Problem 11: A photon of wavelength 5000 A strikes a metal surface, the work function of the metal being 2.20 eV. Calculate (i) the energy of the photon in eV (ii) the kinetic energy of the emitted photo electron and (iii) the velocity of the photo electron.
Solution: i) Energy of the photon
E = hn = = = 3.96 ´ 10–19 J
1 eV = 1.6 ´ 10–19 J
Therefore E = = 2.475 eV
- ii) Kinetic energy of the emitted photo electron
Work function = 2.20 eV
Therefore, KE = 2.475 – 2.20 = 0.275 eV = 4.4 ´ 10–20 J
iii) Velocity of the photo electron
KE = = 4.4 ´ 10–20 J
Therefore, velocity (v) = = 3.11 ´ 105 ms–1
Problem 12: The second ionization potential of Be is 17.98 eV. If the electron in Be+ is assumed to move in a spherical orbit with a central field of effective nuclear charge (Zeff) consisting of the nucleus and other electrons, by how many units of charge is the nucleus shielded by other electrons? (The energy of electron in first Bohr orbit of H is –13.6 eV). If the extent of shielding by the K electrons of Li+ ion is same as you have calculated above, find the ionization potential of Li+?
Solution: Ionization Energy = = 17.98
\ = = 5.28
Zeff = 2.3
Shielding effect = Z–Zeff = 4–2.3 = 1.7
Zeff for Lithium = 3–1.7 = 1.3
\ IE of lithium = = 5.74 eV
Problem 13: Wavelength of the Ka characteristic X–rays of iron and potassium are 1.931´10–8 and 3.737 ´ 10–8 cm respectively. What is the atomic number and name of the element for which characteristic Ka wavelength is
2.289 ´10–8 cm?
At. No of K = 19, and Fe = 26
Velocity of light = 3 ´1010 cm / sec
According to Mosely’s law, the frequency of emitted X – ray n µ Z2
Solution: Frequency n = µ Z2 = K(Z)2
… (1)
… (2)
l1 (Fe) = 1.931 ´ 10–8, Z1 = 26
l3(K) = 3.737 ´10–8, Z3 = 19
l2 (u) = 2.289 ´ 10–8, Z2 = ?
from eqn (1)
On solving we get Z2 = 23.88 = 24
From equation (2)
=
Z2 = 24.27 = 24
Thus, the atomic number of that element is 24 and it is chromium.
Problem 14: The IP of H is 13.6 eV. It is exposed to electromagnetic waves of 1028 A and given out induced radiations. Find the wavelength of these induced radiations?
Solution. E1 of H atoms = –13.6 eV
Energy give to H atom =
= 1.933 ´ 10–18 J = 12.07 eV
\ Energy of H atom after excitation = –13.6 + 12.07
= –1.53 eV
\ En =
\ n2 = = 9
\ n = 3
Thus electron in H atoms is excited to 3rd shell
\ I induced l1 =
Q E1 = -13.6 eV; E3 = –153 eV
\ l1 =
= 1028 ´ 10–10 m
\ l1 = 1028 Å
\ II induced l2 =
Q E1 = –13.6 eV; E2 = – eV
\ l2 = = 1216 ´ 10–10 m = 1215 Å
\ III induced l3 =
Q E1 = –13.6eV; E2 = eV; E3 = eV
\ l3 = = 6568 ´ 10–10 m = 6568 Å
14.2 Objective
Problem 1: If the radius of first Bohr orbit is x, then de Broglie wavelength of electron in 3rd orbit is nearly.
(A) 2px (B) 6px
(C) 9x (D)
Solution: rn = , \ r0 a n2
; r3 = 9r1 = 9x
Also mvrn =
mvr3 =
mv =
l= = 6px,
\ B
Problem 2: When photon of energy 4.25 eV strike the surface of a metal A, the ejected electrons have maximum kinetic energy TA (expressed in eV) and de-Broglie wavelength lA. The maximum kinetic energy of photo electrons liberated from another metal B by photons of energy 4.20 eV is TB = TA – 1.50 eV. The de-Broglie wavelength of these photoelectrons is lB = 2lA then which of the following is incorrect?
(A) The work function of A is 2.25 eV
(B) The work function of B is 3.70 eV
(C) TA = 2.00 eV
(D) TB = 2.75 eV
Solution: = 2 (since lB = 2lA)
= ; Also TA – TB = 1.50
\ TB = 0.50 eV
\ TA = 2.00 eV
Also 4.25 = F0 + TA \FA = 4.25 – 2.00 = 2.25 eV
4.20 = FB + TB ; FB = 4.20 – 0.50 = 3.70 eV
\ (D)
Problem – 3: The radiation is emitted when a hydrogen atom goes from a high energy state to a lower energy state. The wavelength of one line an 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
Solution: DE = hn = = = 3 ´ 10–17 J
\ A
Problem 4: The ratio of the electron in second excited state of He+ ion to the electron in the first excited state of Be3+ is
(A) 1:3 (B) 9:16
(C) 1:9 (D) 16:9
Solution: En = eV
2nd excited state of He+ n = 3, Z = 2 for He+
E3 = =
1st excited state of Be3+ is n = 2 and Z = 4 for Be3+
E2 = =
\ (C)
Problem 5: The number of d-electrons in Fe2+ (Z = 26) is not equal to that of
(A) p-electrons in Ne (Z = 10) (B) s-electrons in Mg (Z = 12)
(C) d-electrons in Fe(Z = 26) (D) p-electrons in Cl (Z = 17)
Solution: Fe2+ (z = 26) electron = 24 [Ar] 3d6, six d-electron
- a) Ne (Z = 10) 1s2 2s2 2p6, six electron in p orbital
- b) Mg (Z = 12) 1s2 2s2 2p6 3s2, six electron in s-orbital
- c) Fe (Z = 26) [Ar] 3d6, six d-electron
\ D
Problem 6: The accelerating potential to be imparted to a beam of electrons to give an effective wavelength of 2.42 nm is
(A) 2.56 V (B) 3.42 V
(C) 0.256 V (D) 0.342 V
Solution: l =
For electron l = × 10–8 cm
2.42 × 10–7 = × 10–8 cm
(24.2)2 =
V = 0.256 Volt
\ (C)
Problem 7: In Bohr’s Model of hydrogen atom the ratio between period of revolution of an electron in the orbit n = 1 to the period of revolution of electron in the orbit n = 2 is
(A) 1:12 (B) 2:1
(C) 1:8 (D) 1:4
Solution: Period of the revolution =
Also
rn = r1 × n2
= T1 : T2 = 1:8
\ (C)
Problem 8: Photoelectric emission is observed from a surface for frequencies n1 and n2 of the incident radiation (n1>n2). If the maximum kinetic energies of the photoelectrons in two cases are in ratio 1:K then the threshold frequency n0 is given by
(A) (B)
(C) (D)
Solution: KE1 = hn1 – hno
KE2 = hn2–hno
It is given that
=
no =
\ (B)
Problem 9: The velocity of electron in the ground state of H atom is 2.184 × 108 cm/sec. The velocity of electron in the first excited state of Li2+ion in cm/sec would be
(A) 3.276 × 108 (B) 2.184 × 108
(C) 4.91 × 108 (D) 1.638 × 108
Solution: vn = (v1H = velocity in the ground state of hydrogen atom)
First excited state is n = 2, Z = 3 for Li+2 ion
vn = = 3.276 × 108
\ (A)
Problem 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
Solution: Energy of electron in first Bohr’s orbit of H–atom
E = (Q ionization energy of H = 2.18 ´10–18J)
E2 = J
= –5.45 ´10–19J
\ (D)
Problem 11. Magnetic moments of V (Z = 23), Cr (Z = 24) and Mn (Z = 25) are x, y, z. Hence
(A) z < y < x (B) x = y = z
(C) x < z < y (D) x < y < z
Solution: Magnetic moments = where n is the number of unpaired electron
V ( Z = 23) (Ar) 3d34s2 n = 3, Ö15 Bm = x
Cr (Z = 24) (Ar) 3d5 4s n = 6, Ö48 Bm = y
Mn (Z = 25) (Ar) 3d5 4s2 n = 5, Ö35 Bm = z
\ C
Problem 12: The speed of a photon is one hundredth of the speed of light in vacuum. What is the de Broglie wavelength? Assume that one mole of protons has a mass equal to one gram. h = 6.626 ´ 10–27 erg sec
(A) 3.31 ´ 10–3Å (B) 1.33 ´ 10–3 Å
(C) 3.13 ´ 10–2 Å (D) 1.31 ´10–2Å
Solution: m =
l = = ´ 6.023 ´ 1023 = 1.33 ´ 10–11 cm
\ B
Problem 13: The wave number of first line of Balmer series of hydrogen atom is 15200 cm–1. What is the wave number of first line of Balmer series of Li2+ ion?
(A) 15200 cm–1 (B) 6080 cm–1
(C) 76000 cm–1 (D) 1,36800 cm–1
Solution: For He+ for H ´Z2 = 15200 ´ 9 = 1, 36,800
\ D
Problem 14: The speed of the electron in the 1st orbit of the hydrogen atom in the ground state is [C is the velocity of light]
(A) (B)
(C) (D)
Solution: V = = 2.189 ´ 108 cm sec–1
C = 3 ´ 1010 cm, = 137
\ D
Problem 15: The quantum number not obtained from the Schrodinger’s wave equation is
(A) n (B) l
(C) m (D) s
Solution: n, l and m quantum numbers can be obtained from Schrodinger equation.
\ D
Problem 16: Among the following the one which represents incorrect set of quantum numbers is
(A) n = 4, l = 2, m = +2 (B) n = 3 , l = 2 , m = 0
(C) n = 3, l = 2, m = –3 (D) n = 4, l = 3, m = + 3
Solution: If n = 4, l = 0, 1, 2, 3
For
l = 0 m = 0
l = 1 m = ± 1, 0
l = 2 m = ± 2, 0 ± 1, 0
l = 3 m = ± 3, +2, ± 1, 0
If n = 3, l = 0, 1, 2
If l = 2 m cannot take a value – 3
\ (C)
Problem 17: Assuming that a 25 watt bulb emits monochromatic yellow light of wave length 0.57 m. 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
Solution: Let n quanta are evolved per sec.
= 25 J sec–1
n = 25
n = 7.18 ´ 1019 sec–1
\ (D)
Problem 18: How many chlorine atoms can you ionize in the process Cl ¾® + e, by the energy liberated from the following process.
Cl + e– ¾® 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
Solution: Energy released in conversion of 6 ´ 1023 atoms of Cl– ions
= 6 ´ 1023 ´ electron affinity = 6 ´ 1023 ´ 3.61 = 2.166 ´ 1024 eV
Let x Cl atoms are converted to Cl+ ion
Energy absorbed = x ´ ionization energy
x ´ 17.422 = 2.166 ´ 1024
x = 1.243 ´ 1023 atoms
\ (A)
Problem 19: The binding energy of an electron in the ground state of the He atom is equal to 24.6 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
Solution : Ionization energy of He
= =
= 54.4 eV
Energy required to remove both the electrons
= binding energy + ionization energy = 24. 6 + 54.4
= 79 eV
\ (C)
Problem 20: The wave number of the shortest wavelength in Lyman series of Li2+ion is
(A) 10.13Å (B) 135.0Å
(C) 13.50Å (D) 101.30Å
Solution: =
= 109678 × 32 = 109678 × 32 cm-1
l = 1.0130 × 10–6 cm
= 101.30 Å
\ (D)
- Assignments (Subjective problems)
Level – I
- Calculate the longest wavelength in the Paschen series of He+ion.
- Calculate the ratio of the wavelength of first and the ultimate line of Balmer series of Li2+.
3 Calculate the number of photons emitted in 10 hours by a 60 W sodium lamp (l = 5893Å).
- An electron collides with a hydrogen atom in its ground state and excited it to a state of n=3. How much energy was given to the hydrogen atom in this inelastic collision?
- Hydrogen atom in states of high quantum number have been created in the laboratory and observed in space.
- find the quantum number of the Bohr orbit in a H–atom whose radius is 0.0100 mm ?
- what is the energy of H–atom in this state?
- With what velocity should an a–particle travel towards the nucleus of a copper atom so as to arrive at a distance 10–13 metre from the nucleus of
the copper atom. [IIT–JEE’97]
- An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated. So that its wave length becomes equal to 1.54 Å. [IIT–JEE ’97]
- If the uncertainty in position and momentum are equal. Calculate the uncertainty in velocity for an electron.
9. Calculate the kinetic energy of an electron emitted from the surface of a metal by light of wavelength 5.5´10–8 cm. Threshold energy for the metal is
2.62´10–9 ergs.
- Suppose 10–17J of energy is needed by the interior of human eye to see an object. How many photons of green light (l=550 nm) are needed to generate this minimum amount of energy?
- When light of frequency 3.2 ´1016 Hz is used to irradiate a metal surface, the maximum kinetic energy of the emitted photoelectrons is 3/4th of the energy of the irradiating photon. What is the threshold frequency of the metal?
- In an atom two electrons move around the nucleus in circular orbits of radii R and 4R. Calculate the ratio of the time taken by them to complete one revolution.
- Show that the de-Broglie’s wavelength of an electron in the nth orbit of a hydrogen atom can be expressed as
l =
- A bulb emits light of wave length 4500Å. 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?
- Calculate the frequency, energy and wave length of radiation corresponding to the spectral line of lowest frequency in Lyman series in the spectra of hydrogen atom. Also calculate the energy for the corresponding line in the spectrum of Li2+ ion. (RH = 1.09678´107m–1, c = 3 ´108 m/s, h = 6.626 ´10–34 Js–1)
Level – II
- The power output for a certain laser transition was found to be 2.79 watt per square meter. Given l = 520 nm, calculate the number of quanta emitted per square meter per second.
- An electron in a hydrogen like atom is in an excited state. It has a total energy of –3.4 eV. Calculate (i) kinetic energy and (ii) de-Broglie wavelength of the electron.
- Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photo electrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find
- The energy of the photons causing the photo electric emission?
- The quantum numbers of the two levels involved in the emission of these photons?
- Find the threshold wavelengths for photoelectric effect from a copper surface, a sodium surface and a caesium surface? The work function of these metals are 4.5 eV, 2.3 eV and 1.9 eV respectively.
- 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 Helium and an atom of oxygen moving with same velocity (h = 6.63 × 10––34 kg m2 sec–1).
- The series limit for the Paschen series of hydrogen spectrum occurs at 8205.8Å. Calculate.
(a) Ionization energy of hydrogen atom
(b) Wave length of the photon that would remove the electron in the ground state of the hydrogen atom.
- The angular momentum of an electron in a particular orbit of Li2+ion is 5.2728 × 10–34 kg m2/sec. Calculate the frequency of the spectral line when electron falls from this level to the level where angular momentum of electron will be 3.1636 × 10–34 kg – m2/sec.
- The electron energy in hydrogen atom is given by E = 21.7 ´ 10–12/ n2 ergs. 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?
- The wavelength corresponding to a transition when electron falls from a certain quantum level to the ground state of an He+ ion is 24.31 nm. Find the ratio of velocity of the electron in the next quantum level to that of velocity of light?
- Calculate the energy emitted when electron of 1.0 g atom of hydrogen undergo transition giving the spectral lines of lowest energy in u.v. region of its atomic spectrum.
(RH = 1.1 ´ 107 m–1, c = 3 ´ 108 ms–1, h = 6.62 ´ 10–34 J–sec)
- When a metal surface is irradiated by light of wave length 300 mm; the stopping potential is found to be 0.5V. Compute the work function and threshold wave length. Also calculate the stopping potential required for light of wave length 200mm.
- The photoelectric effect consists of the emission of electron from the surface of the metal when the metal is irradiated with light. A photon with a minimum energy of 3.97 ´10–19J is necessary to eject an electron from barium.
- What is frequency of the radiation corresponding to this value?
- Will the blue light with wave length 450 nm be able to eject the electron?
- In photoelectric effect, an absorbed quantum of light results in the ejection of an electron from the observer. The kinetic energy of the electron is equal to the energy of the absorbed photon minus the energy of the longest wavelength that causes the effect. Calculate the kinetic energy of an electron produced in cesium by 400 nm light. The critical (maximum) wavelength for the photoelectric effect is cesium is 660 nm.
- Energy required to stop the ejection of electron from Cu plate is 0.24 eV. Calculate the work function when radiation of l= 253.7 nm strikes the plate.
- The dissociation of I2 2I utilizes one photon per iodine molecule dissociated. The maximum l for this is 4995 Å. Calculate number of moles of I2 dissociated per KJ of photon energy.
LEVEL – III
- 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 is first Bohr orbit?
- Electrons in H-like atom (z = 3) make transition from the fifth to the fourth orbit and fourth to third orbit. The resulting radiations are incident on a metal surface and eject photoelectrons. The stopping potential of the photoelectrons ejected by the shorter wavelength is 3.95V. Calculate the work function of the metal and stopping potential for the photoelectrons ejected by the larger wavelength (R = 1.094 × 107/m).
- A hydrogen like atom (atomic no. Z) is in a higher excited state of quantum number ‘n’. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the values of n and Z.
- If the radius of first orbit of He+ ion can be determined with an uncertainty of 1% of its actual value. What will be uncertainty in the velocity of electron? Compare the value with the velocity of electron in the first orbit. (Me = 9.1 × 10–31 kg and = 9.0 × 109 NC–2 m2)
- The energy levels of hypothetical one electron atom are show in the figure.
0 eV ¾¾¾ n =
– 0.50 eV ¾¾¾ n = 5
–1.45eV ¾¾¾ n = 4
– 3.08 eV ¾¾¾ n = 3
– 5.3eV ¾¾¾ n = 2
– 15.6 eV ¾¾¾ n = 1
- a) Find the ionisation potential of atom?
- b) Find the short wavelength limit of the series terminating at n = 2?
- c) Find the wave no. of photon emitted for the transition made by the electron from third orbit to first orbit?
- d) Find the minimum energy that an electron will have after interacting with this atom in the ground sate, if the initial kinetic energy of the electron is (i) 6eV; (ii) 11eV?
- 1.8 g hydrogen atoms are excited to radiations. The study of spectra indciates 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 photo electric emission requires a threshold frequency n0. For a certain metal l1 = 2200Å and l2 = 1900 Å, produce electrons with a maximum kinetic energy KE1 & KE2. If KE2 = 2KE1 calculate n0 and corresponding l0.
- The ionisation energy of a H like Bohr atom is 4 Rydberg.
- a) Calculate the wavelength radiated when electron jumps from the first excited state to ground state.
- b) What is radius of I orbit of this atom?
- Iodine molecule dissociates into atoms after absorbing light of 4500Å. If one quantum of radiation is absorbed by each molecule, calculate the kinetic energy of iodine atoms. Bond energy of I2 = 240 kJ mol–1.
- Two hydrogen atoms collide head on and end up with zero kinetic energy. Each then emits a photon with a wavelength 121.6 nm. Which transition leads to this wavelength? How fast were the hydrogen atoms travelling before the collision? Given RH = 1.097 ´ 107 ms–1 & mH = 1.67 ´ 10–27 kg.
- A mixture contains atoms of fluorine and chlorine. The removal of an electron from each atom of the sample absorbs 272.2 kJ while the addition of an electron to each atom of the mixture releases 68.4 kJ. Determine the percentage composition of the mixture, given that the ionisation energies of F & Cl are 27.91 ´ 10–22 & 20.77 ´ 10–22 kJ respectively and that the electron affinities are 5.53 ´ 10–22 and 5.78 ´ 10–22 KJ respectively.
- A particle of charge equal to that of an electron and mass 200 times the mass of an electron moves in a circular orbital around a nucleus of charge +3e. Assuming that Bohr model of the atom is applicable to this system.
- a) Derive an expression for radius of the nth Bohr orbit.
- b) Find the value of ‘n’ for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom?
- c) Find the wavelength of the radiation emitted when the revolving particle jumps from the third orbit to the first? Given RH = 1.097 ´ 10–7 m–1
- 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?.
- The minimum energy necessary to overcome the attractive force between the electron and the surface of silver metal is 7.52 ´ 10–19 J. What will be the maximum kinetic energy of the electrons ejected from silver which is being irradiated with ultraviolet light having a wavelength 360Å?
- A certain dye absorbs light of l = 5320Å and fluoresces light of wavelength of 7280Å. Assuming that under the given conditions 58% of energy is re-emitted as fluorescence, calculate ratio of number of quanta absorbed to number of quanta emitted out.
- Assignments (Objective problems)
Level – I
- If the electronic structure of oxygen atom is written as 1s2 2s2 2p4 it would violate
(A) Hund’s rule (B) Paulis exclusion principle
(C) Both Hunds’ and Paulis’ principles (D) None of these
- Number of nodal planes for f-orbital are
(A) 3 (B) 2
(C) 1 (D) 0
- One Bohr magneton equals
(A) (B)
(C) (D)
- Three isotopes of an element have mass numbers M, (M + 1) and (M + 2). If the mean mass number is (M + 0.5) than which of the following ratio may be accepted for M, (M + 1), (M + 2) in that order.
(A) 1 : 1 : 1 (B) 4 : 1 : 1
(C) 3 : 2 : 1 (D) 2 : 1 : 1
- A possible set of quantum numbers for the last electron added to a gallium atom
(Z = 31) in its ground state is
n l m mS
(A) 4 1 –1 +1/2
(B) 4 0 0 –1/2
(C) 3 2 +2 +1/2
(D) 3 0 0 –1/2
- The angle made by angular momentum vector of an electron with z axis is given by
(A) cosq = l/m (B) cos = Öl/m
(C) cos q = (D) cosq =
- The distance between 3rd and 2nd orbit of hydrogen atom is
(A) 2.646 ´10–8 cm (B) 2.116 ´10–8 cm
(C) 1.058 ´10–8 cm (D) 0.529 ´ 10–8 cm
- An atom has four unpaired electrons. The total spin of this atom will be
(A) 1 (B) 1.5
(C) 2 (D) 4
- For a ‘d’ electron, the orbital angular momentum is
(A) (B)
(C) (D)
- Which of the following pairs can form correct set of isosters ?
(A) MgS, CaF2 (B) C6H6, B3N3H6
(C) CO, N2O (D) All the above
- The number of electrons present in NO3– ion is
(A) 31 (B) 32
(C) 34 (D) 35
- 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
- Which of the following electronic transitions from one orbit to another corresponds to the third line in the Balmer series of hydrogen spectrum?
(A) 1 ¾®2 (B) 3 ¾®2
(C) 5 ¾®2 (D) ¥ ¾® 2
- The wave number of a spectral line is 5 ´105 m–1. The energy corresponding to this line will be
(A) 3.39´10–23kJ (B) 9.93 ´10–23kJ
(C) 3.45 ´10–24J (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) 1215.6Å (B) 1397.5Å
(C) 2395.87 Å (D) none of these
- The species contains
(A) 25 protons, 25 neutrons, 25 electrons
(B) 23 protons, 30 neutrons, 23 electrons
(C) 25 protons, 30 neutrons, 25 electrons
(D) 25 protons, 30 neutrons, 23 electrons
- The ratio of radii of electron in first Bohr’s orbit of H, He+ and Li2+ is
(A) 1:2: 3 (B) 1:4:9
(C) 3:2:1 (D) 6:3:2
- Which of the following characteristics, a metal should have to show
photo electric effect?
(A) Low ionization potential (B) High ionization potential
(C) Low Electron affinity (D) High electron affinity
- For series limit of a line in hydrogen spectrum, the value of n2 should be
(A) 0 (B) ¥
(C) It depends upon the nature of series (D) None of these
- The amount of energy required to remove an electron from a Li2+ ion in its ground state is how many times greater than the amount of energy needed to remove the electron from a H atom in its ground state?
(A) 9 (B) 6
(C) 4 (D) 3
Level – II
- Electromagnetic radiation (Photon) with least wavelength result when an electron in the hydrogen atom falls from n = 5 to
(A) n = 1 (B) n = 2
(C) n = 3 (D) n = 4
- The first five ionization energies of an element are 801, 2428, 3660, 25030, 32835 in kJ/mol. Then the element could be
(A) a halogen (B) a noble gas
(C) a third group element (D) a second group element
- The energy required to melt 1g ice is 33J. The number of quanta of radiation of frequency 4.67 ´1013 sec–1, that must be absorbed to melt 10 g ice is
(A) 1.065 ´1022 (B) 3.205 ´1023
(C) 9.076 ´1020 (D) none of these
- If the wave length of first line of the Balmer series of hydrogen atom is 656.1 nm, the wave length of second line of this series would be
(A) 218.7 nm (B) 328.0 nm
(C) 486.0 nm (D) 640.0 nm
- The energy of an electron in the first Bohr orbit for He+ ion is –54.4 eV. Which one of the following is a possible excited state for electron in Bohr orbit of He+ ion?
(A) –6.04 eV (B) – 6.8eV
(C) – 1.7 eV (D) +1.36 eV
- The outermost electronic configuration of the most electronegative element is
(A) ns2,np3 (B) ns2,np4
(C) ns2,np5 (D) ns2,np6
- 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
- Ground state electronic configuration of nitrogen atom can be represented by
| (A) | (B) | |||
| (C) | (D) |
- A atom forms an ion by the loss of three electrons. The ion has an electronic configuration [Ar]3d6. The symbol of the ion is
(A) Fe3+ (B) Ni3+
(C) Co3+ (D) Mn+3
- Given that in H–atom, the transition energy for n=1 to n=2 is 10.2 eV, the energy for the same transition in Be3+ is
(A) 20.4 eV (B) 30.6 eV
(C) 40.8 eV (D) None of these
- Which one of the following species is isoelectronic with P3–?
(A) Kr (B) Ca2+
(C) Na+ (D) F–
- Which salt is composed of an isolectronic cation and anion?
(A) KF (B) NaCl
(C) SrCl2 (D) MgF2
- Which of the following violates Pauli’s exclusive principle?
| (A) | (B) | ||
| (C) | (D) |
- The following transition occurs when Lithium atoms are sprayed into hot flame. The various steps are numbered for identification.
2s 2p 3d 3p 4s 3p
Which of these transitions result in emission of light?
(A) 1, 2 & 5 (B) 3 & 5
(C) 3, 4 & 5 (D) All
- The important principles which will help in assigning the electronic configuration of atoms are
(A) (n+l) rule (B) Hund’s rule
(C) Heisenberg’s uncertainity principle (D) Pauli’s exclusion principle
- 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+
- The wavelength of a track star running 150 metre dash in 12 sec if its weight is 50 kg is.
(A) 9.11 ´ 10–34 m (B) 8.92 ´ 10–37 m
(C) 2.15 ´ 10–37 m (D) 1.060 ×10-36 m
- What will be the wavelength of radiation emitted for the electronic transition from infinity to ground state of the He+ ion?
(A) 2.27 ´ 10–8 m (B) 9.12 ´ 10–8 m
(C) 1.01´ 10–8 m (D) none of these
- A proton has wavelength 1Å. The potential by which the proton is accelerated will be
(A) 0.0926 V (B) 0.0502 V
(C) 0.0826 V (D) 51.2 V
- Bond energy of Br2 is 194 kJ mole–1. The minimum wave number of photons required to break this bond is
(A) 1.458 × 1023 m–1 (B) 1.620 × 106 m–1
(C) 4.86 × 1014 m–1 (D) 1.45 × 106 m–1
- Answers to Subjective Assignments
LEVEL – I
- 4689Å 2. 9/5
- 6.4 × 1024 4. 12.08 eV
- a) 435 6. 6.3 × 106m/sec
- b) –7.18 ´ 10–5 ev
- 63.40 8. 7.98 ×1012 m/sec
- 9.93 ´ 10–10 ergs 10. 28
- 8 × 1015 12. 1 : 8
- n = 27.2 ´ 1018
- l = 1215.6 Å, n = 2.4679 ´ 1015 H2
E = 10.2 eV
Energy for corresponding line of Li2+ = 91.80 eV
LEVEL – II
- 7.30 × 1018 2. i) 3.4 eV
- ii) 654Å
- a) 2.55 eV b) 2 ¬ 4
- For Cu = 276 nm
For Mg = 540 nm
For Cs = 654 nm
- a) 1.46 × 10–9 m
- b) l for He is 4 times greater than the l of oxygen atom
- a). I.P. = 13.6 eV 7. 2.1058 × 1015
- b) l = 916Å
- 5.42 ´ 10–12 ergs, 3.67 ´ 10–5 cm 9. 1:344
- 978.75 kJ mole–1 11. 3.63 eV, V0 = 2.57 volt
- a) 5.99 ´ 1014 sec–1 13. 1.22 eV
- b) 417 ´ 10–19 J/atom
- 4.65 eV 15. 4.175 ´ 10–3 moles of I2
LEVEL – III
- a) z = 3 2. 2eV; 0.75V
- b) 8 eV
- c) 01 × 10–8 m
- d) 4 eV
- n = 6, z = 3 4. DV = 2.18 × 108 m/sec
V = 4.36 × 106 m/sec
- a) 15.6 eV 6. i) III Shell = 292. 68 × 1021 atoms
- b) 9nm II Shell = 162.6 × 1021 atoms
- c) 009 × 107m–1 ii) 832.50 KJ
- d) i) Electron will not interact
- ii) 7eV
- 2612.6 8. a) l = 303. 89
- b) 2.645 ×10-9 cm
- 0.216 × 10-19 J 10. 4.43 × 104 m/sec
- % F = 27.27 12. b) 24
% Cl = 72.73 c) 57.36 pm
- 1741.5 Å 14. 47.68 × 10-19 J
- 1.25 : 1
- Answers to Objective Assignments
Level – I
- D 2. A
- D 4. B
- A 6. D
- A 8. C
- A 10. B
- B 12. C
- C 14. B
- A 16. D
- D 18. A
- B 20. A
Level – II
- A 2. C
- A 4. C
- A 6. C
- C 8. A, D
- C 10. D
- B 12. D
- D 14. B
- A, B, D 16. A
- D 18. A 19. C 20. B
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