Chapter 22 Atoms free study material by TEACHING CARE online tuition and coaching classes
Important Atomic Models.
(1) Thomson’s model
J.J. Thomson gave the first idea regarding structure of atom. According to this model.
(i) An atom is a solid sphere in which entire and positive charge and it’s mass is uniformly distributed and in which negative charge (i.e. electron) are embedded like seeds in watermelon.
Success and failure
Explained successfully the phenomenon of thermionic emission, photoelectric emission and ionization.
The model fail to explain the scattering of a– particles and it cannot explain the origin of spectral lines observed in the spectrum of hydrogen and other atoms.
(2) Rutherford’s model
Rutherford’s aparticle scattering experiment
Rutherford performed experiments on the scattering of alpha particles by extremely thin gold foils and made the following observations
Number of scattered particles :
 Most of the aparticles pass through the foil straight away
 Some of them are deflected through small
 A few aparticles (1 in 1000) are deflected through the angle more than 90^{o}.
 A few a particles (very few) returned back e. deflected by 180^{o}.
 Distance of closest approach (Nuclear dimension)
The minimum distance from the nucleus up to which the aparticle approach, is called the distance of closest
approach (r ). From figure r = 1 . Ze2 ; E = 1 mv^{2} =
K.E. of aparticle

0 0 4pe E 2
 Impact parameter (b) : The perpendicular distance of the velocity vector ( v ) of the aparticle from the centre of the nucleus when it is far away from the nucleus is known as impact It is given as
b = Ze ^{2} cot(q / 2)
Þ b µ cot(q / 2)
4pe
æ 1 mv2 ö
_{0} ç 2 ÷


Note : @If t is the thickness of the foil and N is the number of aparticles scattered in a particular direction
(q = constant), it was observed that
N = constant Þ
t
N1 = t1 .
N 2 t 2
After Rutherford’s scattering of aparticles experiment, following conclusions were made as regard as atomic structure :
 Most of the mass and all of the charge of an atom concentrated in a very
small region is called atomic nucleus.
 Nucleus is positively charged and it’s size is of the order of 10^{–15} m » 1 Fermi.
 In an atom there is maximum empty space and the electrons revolve around the nucleus in the same way as the planets revolve around the
10–10 m
leus
Size of the nucleus = 1 Fermi = 10^{–15} m
Size of the atom 1 Å = 10^{–10} m
Draw backs
 Stability of atom : It could not explain stability of atom because according to classical electrodynamic theory an accelerated charged particle should continuously radiate Thus an electron moving in an circular path around the nucleus should also radiate energy and thus move into smaller and
smaller orbits of gradually decreasing radius and it should ultimately fall into nucleus.
 According to this model the spectrum of atom must be continuous where as practically it is a line
 It did not explain the distribution of electrons outside the
(3) Bohr’s model
Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom)
Bohr’s model is based on the following postulates.
 The electron can revolve only in certain discrete nonradiating orbits, called stationary orbits, for which total
angular momentum of the revolving electrons is an integral multiple of h (= h)
2p

i.e. L = n æ h ö = mvr;
2p
where n = 1, 2, 3,……. = Principal quantum number
è ø
 The radiation of energy occurs only when an electron jumps from one permitted orbit to
When electron jumps from higher energy orbit (E_{1}) to lower energy orbit (E_{2}) then difference of energies of these orbits i.e. E_{1} – E_{2} emits in the form of photon. But if electron goes from E_{2} to E_{1} it absorbs the same amount of
energy.
E_{1}
E_{2}
Emission
E_{1} – E_{2} = hn
E_{1}
E_{1} – E_{2} = hn
E_{2}
Absorption
Note : @According to Bohr theory the momentum of an e ^{–}
revolving in second orbit of H atom will be h

p
@ For an electron in the n^{th} orbit of hydrogen atom in Bohr model, circumference of orbit
= nl ; where l = deBroglie wavelength.
Bohr’s Orbits (For Hydrogen and H_{2}Like Atoms).
(1) Radius of orbit
For an electron around a stationary nucleus the electrostatics force of attraction provides the necessary centripetal force
i.e. 1 (Ze)e = mv^{2}
……. (i) also mvr = nh
…….(ii)
4pe _{0} r ^{2} r 2p



From equation (i) and (ii) radius of n^{th} orbit
rn =
n2h2
2
n^{2}h^{2}e n^{2}
2 = ^{0} = 0.53 Å
éwhere k = 1 ù
4p kZme
pmZe^{2} Z
ë 4pe _{0} û



Þ r µ n
Z
Note : @The radius of the innermost orbit (n = 1) hydrogen atom (z = 1) is called Bohr’s radius a_{0} i.e.
a_{0} = 0.53 Å .
(2) Speed of electron
From the above relations, speed of electron in n^{th} orbit can be calculated as
2pkZe ^{2}
Ze ^{2}
æ c ö Z _{6} Z ^{v}
vn =
=
nh 2e
nh = ç
÷. = 2.2 ´ 10
n m / sec
_{0} è 137 ø n
where (c = speed of light 3 ´ 10^{8} m/s)
n
Note : @The ratio of speed of an electron in ground state in Bohr’s first orbit of hydrogen atom to velocity of
light in air is equal to
e ^{2} = 1
(where c = speed of light in air)
(3) Some other quantities
2e _{0} ch 137
For the revolution of electron in n^{th} orbit, some other quantities are given in the following table
(4) Energy
 Potential energy : An electron possesses some potential energy because it is found in the field of nucleus
potential energy of electron in n^{th} orbit of radius r
is given by U = k. (Ze)(e) = – kZe 2

rn rn
 Kinetic energy : Electron posses kinetic energy because of it’s Closer orbits have greater kinetic energy than outer ones.
As we know
mv^{2}
= k.(Ze)(e)
Þ Kinetic energy
K = kZe ^{2}
= U 

r_{n} r ^{2} 2r_{n} 2
 Total energy : Total energy (E) is the sum of potential energy and kinetic energy e. E = K + U
Þ E = –
kZe ^{2}
also rn =
n^{2}h^{2}e _{0}
2
. Hence
æ

E = – ç
me ^{4} ö z ^{2}




2 2 2
æ

= – ç
me ^{4}
2
ö 2


÷ ch
= –Rch Z 2
= 13.6
Z 2 eV
2rn
pmze
ç 8e _{0} h ÷
ç 8e _{0} ch ÷ n n n
where
R = me ^{4}

8e ^{2}ch^{3}
= Rydberg’s constant = 1.09 ´ 10^{7} per metre
Note : @Each Bohr orbit has a definite energy
@ For hydrogen atom (Z = 1) Þ
E = – 13.6 eV
n n2
@ The state with n = 1 has the lowest (most negative) energy. For hydrogen atom it is E_{1} = – 13.6 eV.
@ Rch = Rydberg’s energy ~– 2.17 ´ 10 ^{18} J ~– 31.6 eV .
@ E = –K = U .
2
 Ionisation energy and potential : The energy required to ionise an atom is called ionisation It is the energy required to make the electron jump from the present orbit to the infinite orbit.
æ Z ^{2} ö 13.6Z ^{2}
Hence Eionisation = E¥ – En = 0 – ç 13.6 n2 ÷ = + n2 eV
For H_{2}atom in the ground state
è
Eionisation
ø
= + 13.6(1)^{2} = 13.6 eV
n2
The potential through which an electron need to be accelerated so that it acquires energy equal to the
ionisation energy is called ionisation potential.
Vionisation
= Eionisation
e
 Excitation energy and potential : When the electron is given energy from external source, it jumps to higher energy This phenomenon is called excitation.
The minimum energy required to excite an atom is called excitation energy of the particular excited state and corresponding potential is called exciting potential.
EExcitation
= EFinal
 EInitial
and
VExcitation
= Eexcitation
e
 Binding energy (B.E.) : Binding energy of a system is defined as the energy released when it’s constituents are brought from infinity to form the system. It may also be defined as the energy needed to separate it’s constituents to large If an electron and a proton are initially at rest and brought from large distances to form a hydrogen atom, 13.6 eV energy will be released. The binding energy of a hydrogen atom is therefore 13.6 eV.
Note : @For hydrogen atom principle quantum number n = .
(5) Energy level diagram
The diagrammatic description of the energy of the electron in different orbits around the nucleus is called energy level diagram.
Energy level diagram of hydrogen/hydrogen like atom
n = ¥  Infinite  Infinite  E_{¥} = 0 eV  0 eV  0 eV  
n = 4  Fourth  Third  E_{4} = – 0.85 eV  – 0.85 Z^{2}  + 0.85 eV  
n = 3  Third  Second  E_{3} = – 1.51 eV  – 1.51 Z^{2}  + 1.51 eV  
n = 2  Second  First  E_{2} = – 3.4 eV  – 3.4 Z^{2}  + 3.4 eV  
n = 1  First  Ground  E_{1} = – 13.6 eV  – 13.6 Z^{2}  + 13.6 eV  
Principle  Orbit  Excited  Energy for H_{2}  Energy for H_{2}  Ionisation energy  
quantum  state  – atom  – like atom  from this level (for  
number  H_{2} – atom) 
Note : @In hydrogen atom excitation energy to excite electron from ground state to first excited state will be
– 3.4 – (13.6) = 10.2 eV .
and from ground state to second excited state it is [ – 1.51 – (13.6) = 12.09 eV ].
@ In an
H _{2} atom when
e ^{–} makes a transition from an excited state to the ground state it’s
kinetic energy increases while potential and total energy decreases.
(6) Transition of electron
When an electron makes transition from higher energy level having energy E_{2}(n_{2}) to a lower energy level having energy E_{1} (n_{1}) then a photon of frequency n is emitted
(i) Energy of emitted radiation





 Rc h Z ^{2} æ Rch Z ^{2} ö
E_{2}
_{2} æ 1 1 ö
n_{2}
DE,n,l



DE = E2 – E1 = – ç
÷ = 13.6Z ç – ÷
n2 è
n1 ø
ç n1
n2 ÷
Emission
(ii) Frequency of emitted radiation

DE E_{2} – E_{1} _{2} æ 1 1 ö
DE = hn Þ n = =
h
h = Rc Z
ç
ç n2
÷
n2 ÷
(iii)

Wave number/wavelength
è 1 2 ø
1 n 1
2 æ 1


1 ö 13.6Z ^{2} æ 1 1 ö
Wave number is the number of waves in unit length n
= l = c
Þ l = RZ
ç
ç n2
÷
n2 ÷
ç
hc ç n^{2}
÷
n2 ÷
è 1 2 ø è 1 2 ø
 Number of spectral lines : If an electron jumps from higher energy orbit to lower energy orbit it emits raidations with various spectral
If electron falls from orbit n_{2} to n_{1} then the number of spectral lines emitted is given by
N = (n2 – n1 + 1)(n2 – n1 )
E 2
If electron falls from n^{th} orbit to ground state (i.e. n_{2}
= n and n_{1} = 1) then number of spectral lines emitted
N = n(n – 1)
E 2
Note : @Absorption spectrum is obtained only for the transition from lowest energy level to higher energy levels. Hence the number of absorption spectral lines will be (n – 1).
 Recoiling of an atom : Due to the transition of electron, photon is emitted and the atom is recoiled

h _{2} æ 1 1 ö
Recoil momentum of atom = momentum of photon = l
= hRZ
ç
ç n2
÷
n2 ÷
è 1 2 ø
Also recoil energy of atom = p 2
2m
= h2
2ml^{2}
(where m = mass of recoil atom)
(7) Draw backs of Bohr’s atomic model
 It is valid only for one electron atoms, g. : H, He^{+}, Li^{+2}, Na^{+1} etc.
 Orbits were taken as circular but according to Sommerfield these are
 Intensity of spectral lines could not be
 Nucleus was taken as stationary but it also rotates on its own
 It could not be explained the minute structure in spectrum
 This does not explain the Zeeman effect (splitting up of spectral lines in magnetic field) and Stark effect (splitting up in electric field)
 This does not explain the doublets in the spectrum of some of the atoms like sodium (5890Å & 5896Å)
Hydrogen Spectrum and Spectral Series.
When hydrogen atom is excited, it returns to its normal unexcited (or ground state) state by emitting the energy it had absorbed earlier. This energy is given out by the atom in the form of radiations of different wavelengths as the electron jumps down from a higher to a lower orbit. Transition from different orbits cause
different wavelengths, these constitute spectral series which are characteristic of the atom emitting them. When observed through a spectroscope, these radiations are imaged as sharp and straight vertical lines of a single colour.
Spectral series
The spectral lines arising from the transition of electron forms a spectra series.
 Mainly there are five series and each series is named after it’s discover as Lymen series, Balmer series, Paschen series, Bracket series and Pfund
 According to the Bohr’s theory the wavelength of the radiations emitted from hydrogen atom is given by
1 = R é 1 – 1 ù
l ê n^{2} n^{2} ú
ë 1 2 û
where n_{2} = outer orbit (electron jumps from this orbit), n_{1} = inner orbit (electron falls in this orbit)
 First line of the series is called first member, for this line wavelength is maximum (l_{max})
 Last line of the series (n_{2} = ¥) is called series limit, for this line wavelength is minimum (l_{min})
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 number.
Quantum numbers may be defined as a set of four number with the help of which we can get complete information about all the electrons in an atom. It tells us the address of the electron i.e. location, energy, the type of orbital occupied and orientation of that orbital.

 Principal Quantum number (n) : This quantum number determines the main energy level or shell in which the electron is The average distance of the electron from the nucleus and the energy of the electron
depends on it.
En µ n2
and
rn µ n^{2}
(in Hatom)
The principal quantum number takes whole number values, n = 1, 2, 3, 4,….. ¥
(2) Orbital quantum number (l) or azimuthal 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 L =
h (for a particular value of n).
2p
For a given value of n the possible values of l are l = 0, 1, 2, ….. upto (n – 1)
 Magnetic quantum number (m_{l}) : An electron due to it’s 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
The magnetic quantum number determines the number of preferred orientations of the electron present in a subshell.
The angular momentum quantum number m can assume all integral value between – l to +l including zero.
Thus m_{l} can be – 1, 0, + 1 for l = 1. Total values of m_{l} associated with a particular value of l is given by (2l + 1).
 Spin (magnetic) quantum number (m_{s}) : An electron in 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 Therefore for any particular value of magnetic quantum number, spin quantum number can have two
values, i.e.
m = 1
s 2
(Spin up) or
m = – 1
s 2
(Spin down)
This quantum number helps to explain the magnetic properties of the substance.
Electronic Configurations of Atoms.
The distribution of electrons in different orbitals of an atom is called the electronic configuration of the atom.
The filling of electrons in orbitals is governed by the following rules.
(1) Pauli’s exclusion principle
“It states that no two electrons in an atom can have all the four quantum number (n, l, m_{l} and m_{s}) the same.”
It means each quantum state of an electron must have a different set of quantum numbers n, l, m_{l} and m_{s}. This principle sets an upper limit on the number of electrons that can occupy a shell.
N max
in one shell = 2n^{2}; Thus N_{max} in K, L, M, N …. shells are 2, 8, 18, 32,
Note : @ The maximum number of electrons in a subshell with orbital quantum number l is 2(2l + 1).
(2) Aufbau principle
Electrons enter the orbitals of lowest energy first.
As a general rule, a new electron enters an empty orbital for which (n + l ) is minimum. In case the value
(n + l) is equal for two orbitals, the one with lower value of n is filled first.
Thus the electrons are filled in subshells in the following order (memorize) 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, ……
(3) Hund’s Rule
When electrons are added to a subshell where more than one orbital of the same energy is available, their spins remain parallel. They occupy different orbitals until each one of them has at least one electron. Pairing starts only when all orbitals are filled up.
Pairing takes place only after filling 3, 5 and 7 electrons in p, d and f orbitals, respectively.
Example: 1 The ratio of areas within the electron orbits for the first excited state to the ground state for hydrogen atom is [BCECE 2004]
(a) 16 : 1 (b) 18 : 1 (c) 4 : 1 (d) 2 : 1
Solution : (a) For a hydrogen atom

2 r 2 n4
pr ^{2} n^{4}
A n^{4} 2^{4}
A 16
Radius r µ n
Þ 1 = 1 Þ 1 = 1
Þ 1 = 1 =
= 16
Þ 1 =





2 4 2 4
2 2 2
2 4 14
A2 1


Example: 2 The electric potential between a proton and an electron is given by
V = V0
ln r , where
r0
r_{0} is a constant.
Assuming Bohr’s model to be applicable, write variation of rn with n, n being the principal quantum number
[IITJEE (Screening) 2003]
(a)
rn µ n
(b)
rn µ 1 / n
(c)
rn µ n^{2}
rn µ 1 / n^{2}
Solution : (a) Potential energy U = eV = eV0 ln r
r0
\ Force
F = – dU
dr
= eV0 . The force will provide the necessary centripetal force. Hence
r
mv ^{2} = eV_{0}
Þ v =
…..(i) and
mvr = nh
…..(ii)
r r 2p
Dividing equation (ii) by (i) we have
mr = æ nh ö

2p
or r µ n
è ø
Example: 3 The innermost orbit of the hydrogen atom has a diameter 1.06 Å. The diameter of tenth orbit is
(a) 5.3 Å (b) 10.6 Å (c) 53 Å (d) 106 Å
[UPSEAT 2002]
r æ n ö ^{2}
d æ n ö ^{2}
d æ 10 ö ^{2}
Solution : (d) Using r µ n^{2} Þ ^{ } ^{2} = ç ^{ } ^{2} ÷
or ^{2} = ç ^{ } ^{2} ÷
Þ ^{ } ^{2} = ç ÷ Þ d = 106 Å
r1 è n1 ø
d1 è n1 ø
1.06
è 1 ø
Example: 4 Energy of the electron in n^{th} orbit of hydrogen atom is given by En
= – 13.6 eV . The amount of energy needed
n2
to transfer electron from first orbit to third orbit is [MH CET 2002; Kerala PMT 2002]
(a) 13.6 eV (b) 3.4 eV (c) 12.09 eV (d) 1.51 eV
Solution : (c) Using
E = – 13.6 eV
n2
For n = 1 , E1 = –13.6 = 13.6 eV
12
and for n = 3 E3 = – 13.6 = 1.51 eV
32
So required energy = E_{3} – E_{1} = 1.51 – (13.6) = 12.09 eV
Example: 5 If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the
electron from the first excited state of Li ^{+} ^{+} is [AIEEE 2003]
(a) 122.4 eV (b) 30.6 eV (c) 13.6 eV (d) 3.4 eV
13.6 ´ Z ^{2}
Solution : (b) Using
En = – eV
n2
For first excited state n = 2 and for Li ^{++} , Z = 3
\ E = – 13.6 ´ 3^{2} = – 13.6 ´ 9 = 30.6 eV . Hence, remove the electron from the first excited state of Li ^{++} be 30.6 eV
22 4
Example: 6 The ratio of the wavelengths for 2 ® 1 transition in Li^{++}, He^{+} and H is [UPSEAT 2003]
(a) 1 : 2 : 3 (b) 1 : 4 : 9 (c) 4 : 9 : 36 (d) 3 : 2 : 1

1 _{2} æ 1 1 ö 1
1 1 1
Solution : (c) Using
= RZ ç –
÷ Þ µ

2
2 Þ lLi : lHe+
: l_{H} =
: : = 4 : 9 : 36


l ç n1
n2 ÷ Z
9 4 1
Example: 7 Energy E of a hydrogen atom with principal quantum number n is given by
E = –13.6 eV . The energy of a
n2
Solution : (a)
photon ejected when the electron jumps n = 3 state to n = 2 state of hydrogen is approximately
[CBSE PMT/PDT Screening 2004]

(a) 1.9 eV (b) 1.5 eV (c) 0.85 eV (d) 3.4 eV
= .6 – = 13.6 ´ = 1.9 eV


ç 22 32 ÷ 36
Example: 8 In the Bohr model of the hydrogen atom, let R, v and E represent the radius of the orbit, the speed of electron and the total energy of the electron respectively. Which of the following quantity is proportional to the quantum number n [KCET 2002]
 R/E (b) E/v (c) RE (d) vR
e _{0}n^{2}h^{2}
Solution : (d) Rydberg constant R = pmZe 2
Velocity v =
Ze ^{2}
2e _{0}nh
and energy
E = –
mZ ^{2}e ^{4}

8e ^{2}n^{2}h^{2}
Now, it is clear from above expressions R.v µ n
Example: 9 The energy of hydrogen atom in nth orbit is E_{n}, then the energy in nth orbit of singly ionised helium atom will be [CBSE PMT 2001]
(a) 4E_{n} (b) E_{n}/4 (c) 2E_{n} (d) E_{n}/2
13.6 Z ^{2}
E æ Z ö ^{2}
æ 1 ö ^{2}
Solution : (a) By using
E = –
Þ H = ç H ÷
= ç ÷
Þ EHe = 4 En .
n2 EHe è ZHe ø
è 2 ø
Example: 10 The wavelength of radiation emitted is l_{0}
when an electron jumps from the third to the second orbit of
hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be [SCRA 1998; MP PET 2001]
(a)
16 l (b)
20 l (c)
27 l (d)
25 l
25 ^{0}
27 ^{0}
20 ^{0}
16 ^{0}
Solution : (b) Wavelength of radiation in hydrogen atom is given by
1 = Ré 1 – 1 ù Þ 1 = Ré 1 – 1 ù = Ré 1 – 1 ù = 5 R
…..(i)
l ê n2 n2 ú l
êë 2^{2}
3^{2} úû
êë 4
9 úû 36
ëê 1
_{2} úû 0
and
1 = Ré 1 –
1 ù = Ré 1 –
1 ù = 3R
…..(ii)
l ¢ ëê 22
4 ^{2} úû êë 4
16 úû 16
From equation (i) and (ii)
l ¢ = 5R ´ 16 = 20
Þ l ¢ = 20 l
l 36 3R 27
27 ^{0}
Example: 11 If scattering particles are 56 for 90 ^{o}
angle then this will be at 60 ^{o}
angle [RPMT 2000]
(a) 224 (b) 256 (c) 98 (d) 108
Solution : (a) Using Scattering formula
é æ q
ö ù 4
é æ 90° ö ù 4
ê sinç ^{1} ÷ ú êsinç ÷ ú 4
N µ 1 Þ
N 2 = ê è 2 ø ú
Þ N 2 = ê è 2 ø ú
= é sin 45°ù
= 4 Þ N
= 4 N
= 4 ´ 56 = 224
sin ^{4} (q / 2)
N_{1} ê æ q _{2} ö ú
N_{1} ê æ 60° ö ú
êë sin 30° úû 2 1
ê sinç
2 ÷ ú
êsinç 2 ÷ ú
ëê è
ø úû
ë è ø û
Example: 12 When an electron in hydrogen atom is excited, from its 4^{th} to 5^{th} stationary orbit, the change in angular momentum of electron is (Planck’s constant: h = 6.6 ´ 10 ^{–}^{34} J–s ) [AFMC 2000]
(a)
4.16 ´ 10 ^{–}^{34} J– s
(b)
3.32 ´ 10 ^{–}^{34} J–s
(c)
1.05 ´ 10 ^{–}^{34} J–s
(d)
2.08 ´ 10 ^{–}^{34} J–s
Solution : (c) Change in angular momentum
DL = L
 L = n2 h – n1h
Þ DL = h (n – n ) = 6.6 ´ 10 34 (5 – 4) = 1.05 ´ 10 ^{–}^{34} Js
^{2} ^{1} 2p 2p
2p 2 1
2 ´ 3.14
Example: 13 In hydrogen atom, if the difference in the energy of the electron in n = 2 and n = 3 orbits is E, the ionization energy of hydrogen atom is [EAMCET (Med.) 2000]
(a) 13.2 E (b) 7.2 E (c) 5.6 E (d) 3.2 E
Solution : (b) Energy difference between n = 2 and n = 3;
E = Kæ 1
– 1 ö = Kæ 1 – 1 ö = 5 K
…..(i)

ç 22
32 ÷ ç
÷ 36





Ionization energy of hydrogen atom n
= 1 and n
= ¥ ;
E¢ = Kæ 1 – 1 ö = K
…..(ii)


From equation (i) and (ii)
1 2
E¢ = 36 E = 7.2 E
5
ç 12
¥ 2 ÷
Example: 14 In Bohr model of hydrogen atom, the ratio of periods of revolution of an electron in n = 2 and n = 1 orbits is
[EAMCET (Engg.) 2000]
(a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 16 : 1
3 Þ T2
= n3 = 23 = 8
Solution : (c) According to Bohr model time period of electron T µ n
2
Þ T2 = 8T1 .



1 3 13 1
Example: 15 A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of
required radiation for emitting electron from first to third Bohr orbit in
Li++
will be (Ionisation energy of
hydrogen atom is 13.6 eV) [IITJEE 1985; UPSEAT 1999]
(a) 182.51 Å (b) 177.17 Å (c) 142.25 Å (d) 113.74 Å
Solution : (d) Energy of a electron in nth orbit of a hydrogen like atom is given by
E = 13.6 Z 2 eV , and Z = 3 for Li
n n2
Required energy for said transition
DE = E
– E = 13.6Z ^{2} æ 1 –
1 ö = 13.6 ´ 3^{2} é 8 ù = 108.8 eV = 108.8 ´ 1.6 ´ 10 ^{–}^{19} J
3 1 ç 12
32 ÷
êë 9 úû


Now using DE = hc
l
Þ l = hc
DE
Þ l = 6.6 ´ 10 ^{–}^{34} ´ 3 ´ 10^{8}
108.8 ´ 1.6 ´ 10 ^{–}^{19}
= 0.11374 ´ 10 ^{–}^{7} m Þ l = 113.74 Å
Example: 16 The absorption transition between two energy states of hydrogen atom are 3. The emission transitions between these states will be [MP PET 1999]
(a) 3 (b) 4 (c) 5 (d) 6
Solution : (d) Number of absorption lines = (n – 1) Þ 3 = (n – 1) Þ n = 4
Hence number of emitted lines = n(n – 1) = 4(4 – 1) = 6
2 2
Example: 17 The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E/3 and 2E respectively. A photon of wavelength l is emitted for a transition 3 ® 1. What will be the wavelength of emissions for transition 2 ® 1
[CPMT 1996]
(a) l/3 (b) 4l/3 (c) 3l/4 (d) 3l
Solution : (d) For transition 3 ® 1
DE = 2E – E = hc
l
Þ E = hc
l
…..(i)
For transition 2 ® 1
4 E – E = hc Þ E = 3hc
…..(ii)
From equation (i) and (ii)
3 l ¢ l ¢
l ¢ = 3l
Example: 18 Hydrogen atom emits blue light when it changes from n = 4 energy level to n = 2 level. Which colour of light would the atom emit when it changes from n = 5 level to n = 2 level [KCET 1993]
(a) Red (b) Yellow (c) Green (d) Violet
Solution : (d) In the transition from orbits
5 ® 2
more energy will be liberated as compared to transition from 4 ® 2. So
emitted photon would be of violet light.
Example: 19 A single electron orbits a stationary nucleus of charge +Ze, where Z is a constant. It requires 47.2 eV to excited electron from second Bohr orbit to third Bohr orbit. Find the value of Z [IITJEE 1981]
(a) 2 (b) 5 (c) 3 (d) 4
Solution : (b) Excitation energy of hydrogen like atom for n_{2} ® n_{1}
DE = 13.6Z
Þ Z = 5
ç –

1 2
.2 = 13.6Z ç
Z Þ Z =
= 24.98 ~ 25
Example: 20 The first member of the Paschen series in hydrogen spectrum is of wavelength 18,800 Å. The short wavelength limit of Paschen series is [EAMCET (Med.) 2000]
(a) 1215 Å (b) 6560 Å (c) 8225 Å (d) 12850 Å
Solution : (c) First member of Paschen series mean it’s l_{max} = 144
7R
Short wavelength of Paschen series means l_{min} = 9
R
Hence
lmax = 16 Þ l
= 7 ´ l
= 7 ´ 18,800 = 8225 Å .
lmin 7
min 16
max 16
Example: 21 Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is
[EAMCET (Engg.) 1995; MP PMT 1997]
(a) 1 : 3 (b) 27 : 5 (c) 5 : 27 (d) 4 : 9
Solution : (c) For Lyman series
1 = R é 1 –
1 ù = 3R
…..(i)


l êë1^{2} 2^{2} úû 4
For Balmer series
1 = R é 1
– 1 ù = 5R
…..(ii)


l êë 2^{2}
32 úû 36
From equation (i) and (ii)
lL1 = 5 .

l 27
1
Example: 22 The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be [RPET 1997]
(a) 3.4 eV (b) 13.6 eV (c) 54.4 eV (d) 122.4 eV
1 _{2} æ 1 1 ö 1
_{7} _{2} æ 1 1 ö
Solution : (c) Using
= RZ ç – ÷ Þ
9 = 1.1 ´ 10
´ Z ç 2 – 2 ÷




l ç n1 n2 ÷
108.5 ´ 10
è 2 5 ø
Þ 1
108.5 ´ 10 ^{–}^{9}
= 1.1 ´ 10^{7} ´ Z ^{2} ´
21
100
Þ Z ^{2} =
100 = 4
108.5 ´ 10 ^{–}^{9} ´ 1.1 ´ 10 ^{–}^{7} ´ 21
Þ Z = 2
Now Energy in ground state
E = 13.6Z ^{2} eV = 13.6 ´ 2^{2} eV = 54.4 eV
Example: 23 Hydrogen (H), deuterium (D), singly ionized helium
(He^{+})
and doubly ionized lithium
(Li ^{++} )
all have one
electron around the nucleus. Consider
n = 2 to
n = 1
transition. The wavelengths of emitted radiations are
l_{1}, l_{2}, l_{3} and l_{4} respectively. Then approximately [KCET 1994]
(a)
l_{1} = l_{2} = 4l_{3} = 9l_{4}
(b)
4l_{1} = 2l_{2} = 2l_{3} = l_{4}
(c)
l_{1} = 2l_{2} = 2
2l_{3} = 3
2l_{4} (d)
l_{1} = l_{2} = 2l_{3} = 3l_{4}
Solution : (a) Using DE µ Z ^{2}
(∵ n_{1} and n_{2} are same)
Þ hc µ Z ^{2}
l
Þ lZ ^{2} = constant Þ l_{1}
Z ^{2} = l_{2}
Z ^{2} = l_{3}
Z ^{2} = l_{4}
Z ^{4} Þ l_{1}
´ 1 = l_{2}
´ 1^{2} = l_{3}
´ 2^{2} = l_{4}
´ 3^{3}



Þ l_{1} = l_{2} = 4l_{3} = 9l_{4} .
Example: 24 Hydrogen atom in its ground state is excited by radiation of wavelength 975 Å. How many lines will be there in the emission spectrum [RPMT 2002]
(a) 2 (b) 4 (c) 6 (d) 8
Solution : (c) Using
1 = Ré 1 – 1 ù
Þ 1 = 1.097 ´ 10^{7} æ 1 – 1 ö
Þ n = 4
l ê n2
n2 ú
975 ´ 10 ^{–}^{10}
ç 12
n 2 ÷
ëê 1
_{2} úû è ø
Now number of spectral lines
N = n(n – 1) = 4(4 – 1) = 6 .
2 2
Example: 25 A photon of energy 12.4 eV is completely absorbed by a hydrogen atom initially in the ground state so that it is excited. The quantum number of the excited state is [UPSEAT 2000]
(a) n =1 (b) n= 3 (c) n = 4 (d) n = ¥
Solution : (c) Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation
Þ – 13.6 = 13.6 + 12.4 Þ – 13.6 = 1.2 Þ n ^{2} = 13.6 = 12 Þ n = 3.46 ≃ 4
n 2 n 2
1.2
Example: 26 The wave number of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is
20,397 cm^{–1}. The wave number of the energy for the same transition in He^{+} is [Haryana PMT 2000]
(a) 5,099 cm^{–1} (b) 20,497 cm^{–1} (c) 40,994 cm^{–1} (d) 81,998 cm^{–1}
1 æ 1 1 ö
n æ Z ö ^{2} æ Z ö 2
Solution : (d) Using
= n = RZ ^{2} ç – ÷ Þ n
µ Z ^{2} Þ ^{ } ^{2} = ç 2 ÷ = ç ÷ = 4
Þ n _{2} = n ´ 4 = 81588 cm^{–}^{1} .
l ç n2 n2 ÷
n 1 è Z1 ø è 1 ø
è 1 2 ø
Example: 27 In an atom, the two electrons move round the nucleus in circular orbits of radii R and 4R. the ratio of the time taken by them to complete one revolution is
(a) 1/4 (b) 4/1 (c) 8/1 (d) 1/8
n3
Solution : (d) Time period T µ Z 2
For a given atom (Z = constant) So T µ n^{3} …..(i) and radius R µ n^{2}
…..(ii)
T æ R
ö 3 / 2
æ R ö ^{3} ^{/} ^{2} 1
\ From equation (i) and (ii) T µ R ^{3} ^{/} ^{2} Þ ^{1} = ç 1 ÷ = ç ÷ = .
T2 è R2 ø
è 4 R ø 8
Example: 28 Ionisation energy for hydrogen atom in the ground state is E. What is the ionisation energy of Li^{++} atom in the 2^{nd} excited state
 E (b) 3E (c) 6E (d) 9E
Z 2
Solution : (a) Ionisation energy of atom in nth state En =
n2
For hydrogen atom in ground state (n = 1) and Z = 1 Þ E = E0
…..(i)
For
Li ^{++} atom in 2^{nd} excited state n = 3 and Z = 3, hence
E¢ = E0 ´ 3^{2} = E
32 0
…..(ii)
From equation (i) and (ii) E¢ = E .
Example: 29 An electron jumps from n = 4 to n = 1 state in Hatom. The recoil momentum of Hatom (in eV/C) is (a) 12.75 (b) 6.75 (c) 14.45 (d) 0.85
Solution : (a) The Hatom before the transition was at rest. Therefore from conservation of momentum
Photon momentum = Recoil momentum of Hatom or P
= hn
= E4 – E1
= 0.85eV (13.6 eV) = 12.75 eV
recoil c c c c
Example: 30 If elements with principal quantum number n > 4 were not allowed in nature, the number of possible elements would be
[IITJEE 1983; CBSE PMT 1991, 93; MP PET 1999; RPET 1993, 2001; RPMT 1999, 2003; J & K CET 2004]
(a) 60 (b) 32 (c) 4 (d) 64
Solution : (a) Maximum value of n = 4
So possible (maximum) no. of elements
N = 2 ´ 1^{2} + 2 ´ 2^{2} + 2 ´ 3^{2} + 2 ´ 4 ^{2} = 2 + 8 + 18 + 32 = 60 .
Tricky example: 1 
If the atom 100 Fm^{257} follows the Bohr model and the radius of 100 Fm^{257} is n times the Bohr radius, then find n
[IITJEE (Screening) 2003] (a) 100 (b) 200 (c) 4 (d) ¼
Solution : (d) (r ) = æ m2 ö(0.53 Å) = (n ´ 0.53 Å) Þ m2 = n _{m} ç Z ÷ Z ç ÷ è ø m = 5 for 100 Fm ^{257} (the outermost shell) and z = 100 \ n = (5)^{2} = 1 100 4 
Tricky example: 2 
An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (in eV) required to remove both the electrons from a neutral helium atom is [IITJEE 1995]
(a) 79.0 (b) 51.8 (c) 49.2 (d) 38.2
Solution : (a) After the removal of first electron remaining atom will be hydrogen like atom. So energy required to remove second electron from the atom E = 13.6 ´ 22 = 54.4 eV 1 \ Total energy required = 24.6 + 54.4 = 79 eV 