HINTS & SOLUTIONS

1. (B)
De-broglie wavelength of electron in hydrogen
h 2πrn
Atom = =
mv n
For second Bohr orbit,
600 × 10−9
λ dB = = 300 × 10−9 m …(i)
2
150
λ dB = Å = 3000 Å ...(ii)
V
From (i) and (ii) we get:
150 5
V= 2
= × 10−5 volt
(3000) 3

2. (D)
Angular momentum
n2 h
L = 4.2175 × 10−34 =
2π
⇒ n2 = 4
For the transition from n2 = 4 to n = 3
The wavelength of spectral line λeV
1 13.6  1 1
= −
λ hc  32 42 


13.6eV  7 
=  
1240eVnm  9 × 16 
1240 × 144
λ= = 1876 nm
13.6 × 7
= 18760 Å = 1.876 × 104 Å

3. (A)
For the first of Balmer series of hydrogen
1  1 1  5R 36
= R 2 − 2  = ⇒λ=
λ  2 3  36 5R
For singly ionised helium z = 2
1  1 1 
= 4R  2 − 2 
λ′  
 n2 n1 
Given λ ′ = λ
For n1 = 6 to n2 = 4
1  1 1  20 R
= 4R  2 − 2  =
λ′ 4 6  144
36
λ′ =
5R
It corresponds to transition from n1 = 6 to n2 = 4
4. (C)
1 1 1 
= RZ 2  2 − 2  ;
λ  n1 n2 
 1 1 1 5
= RZ 2  −  = RZ 2  
λ longest Balmer  4 9   36 
1
= RZ 2 [1];
λshortest Lyman
36 1 31
∆λ = 2
− 2
=
5 RZ RZ 5 RZ 2
31
∴ R=
5∆λZ 2

5. (B)
ε0 n 2 h 2 ε0 (2πL)  nh 
r= ⇒ r=  L = 
e2 πm e 2 πm  2π 
⇒ Lr–1/2 = Constant.

6. (C)
1 1 3
First excitation energy = RhC =  2 − 2  = RhC
1 2  4
3
∴ RhC = V e.v.
4
4V
∴ RhC = e.v.
3

7. (B)
In the first case K.E. of H-atom increases due to recoil
whereas in the second case K.E. decreases due to recoil
but E1 + KE1 = E2 + KE2.
E2 > E1

8. (C)
Energy of photon is given by mc2 now the maximum energy of photon is equal to the maximum energy of electron
= eV.
Hence mc2 = eV
eV 1.6 × 10−19 × 18 × 103
⇒ m= 2 = = 3.2 × 10−32 kg
c (3 × 108 )2

9. (D)
1 1 1 
Using = R ( z − 1) 2  2 − 2 
λ  n2 n1 
For a particle: n1 = 2, n2 = 1
1875 R  3
For metal A; = R ( z1 − 1) 2  
4  4
⇒ z1 = 26
 3
For metal B;675 R = R ( z2 − 1) 2  
 4
⇒ z2 = 21

10. (D)
For 2nd line of Balmer series in hydrogen spectrum
1  1 1 3 1  1 1 3
= R(1)  2 − 2  = R For Li 2+ = R(3)2  2 − 2  = R
λ 2 4  16 λ 2 4  16
which is satisfied by only (D).
11. (D)
Energy of nth state in Hydrogen is same as energy of 3nth state in Li++.
3 – 1 transition in H would give same energy as the 3 × 3 ⇒ 1 × 3 transition in Li++.

12. (A)
1 3R
= ( Z − 1) 2
λα 4
4 4
⇒ ( Z − 1) = =
3 Rλ a 3 × 1.1 × 10 × 1.8 × 10−10
7

200 5 78
= = = 26 ⇒ Z = 27
3 33 3

13. (B)
λ m will increase to 3λ m due to decrease in the energy of bombarding electrons. Hence, no characteristics X-rays
will be visible, only continuous X-ray will be produced.

14. (D)
µ0 I e
∵ B= and I =
2rT T
µ e
B = 0 [ r ∝ n 2 , T ∝ n3 ]
2rT
1
B∝ 5
n
15. (C)
For hydrogen atom, we get;
1  1 1  1 3 1 3
= RZ 2  2 − 2  ⇒ = R (1) 2   ⇒ = R (1) 2  
λ n   4  λ2 4
 1 n2  λ1
1  3 1  3
⇒ = R (2) 2   ⇒ = R (3)2  
λ3  4 λ4  4
1 1 1 1
∴ = = =
λ1 4λ 3 9λ 4 λ 2

16. (B)
1 1 1
The wavelength of a spectral line in the Lyman series is = R  2 − 2  , n = 2, 3, 4, … and that in the Balmer
λL 1 n 
1  1 1
series is = R  2 − 2  , n = 3, 4, 5, …
λB 2 n 
For the longest wavelength in the Lyman series, n = 2
1 1 1 1 1  4 − 1 3R 1 4
∴ = R 2 − 2  = R −  = R  = or =
λL 1 2   1 4  4  4 λ L 3R
For the longest wavelength in Balmer series, n = 3
1  1 1  1 1  9 − 4  5R 36
∴ = R 2 − 2  = R −  = R  = or λ B =
λB 2 3   4 9  36  36 5R
λ 4 5R 5
Thus, L = × =
λ B 3R 36 27
17. (D)
Z2 4 4
Energy of electron in He+ 3rd orbit E3 = −13.6 × 2 eV = −13.6 × eV = −13.6 × × 1.6 × 10−19 J = 9.7 × 10−19 J
n 9 9
 As per Bohr’s model,
1
Kinetic energy of electron in the 3rd orbit = –E3 ∴ 9.7 × 10−19 = me v 2
2

2 × 9.7 × 10−19
v= = 1.46 × 106 ms −1
9.1 × 10−31

18. (D)
∆E = hv where, ∆E is energy of radiation, h is Plank’s constant and v is frequency.

∆E  1 1  k 2n 2k 1
⇒v= =k 2
− 2 = 2 2
= 3∝ 3
h  (n − 1) n  n (n − 1) n n

19. (D)
n(n − 1) 4(4 − 1)
In emission spectrum, number of bright lines is given by = =6
2 2

20. (C)
At the distance of closest approach d,
1 2 1 (2e)( Ze)
Kinetic energy = potential energy mv =
2 4πε 0 d
Where,
Ze = charge of target nucleus
2e = charge of alpha nucleus
1 2
mv = kinetic energy of alpha nucleus of mass m moving with velocity v
2
2 Ze 2 1
or d = ∴d ∝
1  m
4πε 0  mv 2 
2 
21. (B)
Energy of emitted photon
1 1 3
E =  2 − 2  × 13.6eV = × 13.6eV
1 2  4
Energy required to completely remove the electron from nth excited state of doubly ionized lithium,
13.6 Z 2 13.6 × 9
E′ = 2
eV = eV
n n2
As E ≥ E ′
3 13.6 × 9
∴ × 13.6 ≥ 2
⇒ n 2 ≥ 3 × 4 or n ≥ 12 = 3.5
4 n
Least quantum number for the excited state = 4
22. (C)
For an electron in nth excited state of hydrogen atom,
e2
kinetic energy =
8πε 0 n 2 a0
 −e2
potential energy =
4πε 0 n 2 a0
−e2
and total energy =
8πε 0 n 2 a0
where a0 is Bohr radius.
As electron makes a transition from an excited state to the ground state, n decreases. Therefore kinetic energy increase
but potential energy and total energy decrease.

23. (B)
nh
mvR = …(i)
2π
mv 2 mv
and qvB = ; qB = …(ii)
R R
From eqns. (i) and (ii), we get
 nh 
qB  = mv
 2πmv 
1 2 1
mv = nhqB
2 4πm
1 2 1  hqB 
mv = nhqB ∴ E = n
2 4πm  4πm 

24. (B)
The accelerating potential difference for the electrons to produce minimum wavelength 66.3 pm is given
12431
As V = volts ≈ 18.75kV
λ
h
The De-Broglie wavelength of the electrons reaching the anode is given as λ = ≈ 8.9pm
2meV
Hence only option (B) is correct
25. (A)
mv 2 3q 2 3q 2
= ⇒ mvr = …(i)
r 4πε 0 r 2 4πε 0 v
nh
and = mvr …(ii)
2π
26. (A)
U = ar2
dU
⇒ F =− = −2ar
dr
Negative sign indicates that force is towards the centre of the circle. It must be centripetal force.
mv 2
∴ = 2ar
r
2a
⇒ v= ⋅r …(i)
m
h
Given: mvr = n
2π
1/4
 2a  nh  n2 h2 
⇒ m ⋅ r r = ⇒ r= 2
 m  2π  8amπ 
27. (C)
m⋅m m
Reduced mass µ = =
m+m 2
 1
∴ Energy values are compared to hydrogen.
2

28. (D)
We assume that orbit is circular (though the electron will spiral down into the nucleus). If orbital radius at an instant
is r then equating the electrostatic force to the centripetal force we get
mv 2 1 e2
= …(i)
r 4π∈0 r 2

e2
and v =
4π ∈0 rm

1 2 1 1 e2 e2
Kinetic energy E = mv = ⋅ = …(ii)
2 2 4π ∈0 r 8π ∈0 r
2πr
Time period of circular motion is T =
v
 dE 
Energy loss in one revolution is ∆E    T
 dt 
2
e2 a 2
2πr e2 r  v 2   v2 
= ⋅ =   ∵ a = 
6π ∈0 c3 v 3 ∈0 c3v  r   r 

e 2 v3
=
3ε 0 c3r

∆E ∆E e 2 v3 8π ∈0 r
= = × [using (ii)]
E 1 2 3∈ c r 3
e2
mv 0
2
3
8π  v 
=  
3  c

29. (A)
dU U 0
U = U 0 nr − U 0 nr0 ⇒ =
dr r
dU U
∴ Force between the particles is F = − =− 0
dr r
Negative sign indicates attraction
mv 2 U 0
∴ = …(i)
r r
nh
And mvr = …(ii)
2π
nh
From (i) and (ii) r =
2π mU 0
1 U
Kinetic energy of electron K = mv 2 = 0 [using (i)]
2 2
 U0 r
Energy of electron in nth orbit is En = KE + PE = + U 0 n  
2  r0 

U0  nh 1
= + U 0 n  . 
2  2π mU 0 r0 

U0   n2 h2  
= 1 + n  2 2

2   4π mU 0 r0  

∆Enm = En − Em

U 0  n2 h2  U 0  m2 h2 
= n  2 2
− n  2 2
2  4π mU 0 r0  2  4π mU 0 r0 
2
 n  n
=  n   = 2 n  
 m  m

 1
n  
∆E12  2
∴ = =1
∆E24  2
n  
 4

30. (D)
13.6(2) 2
E = 24.6 + = 79.0 eV
12

31. (D)
N = nC2 = 5C2 = 10

32. (C)
hc 1242eV − nm
E= = = 59 keV
λ 0.021nm

33. (C)
1 1 1 
= R ( Z − 1) 2  2 − 2 
λ  n1 n2 
For Kα line, n1 = 1, n2 = 2
1 1 1  3
= R (43 − 1) 2  2 − 2  = R (42)2   …(i)
λ 1 2   4

1 1 1  3
and = R(29 − 1) 2  2 − 2  = R(28) 2   …(ii)
λ′ 1 2   4
λ′ 9 9
Dividing equation (i) by (ii), we get = ⇒ λ′ = λ
λ 4 4
34. (B)

1.89 5
z= = 0.185 =
10.2 27
hc hc
= 1.89 = 10.2
λ1 λ2
λ1 10.2
Now = = 5.39
λ 2 1.89
h h P λ P 1.89 5
λ= P= ⇒ 2 = 2 1 = =
p λ P2 λ1 P2 10.2 27

35. (A, C, D)
Given, incident wavelengths travel in x-direction which are in U.V region, which means they can excite an electron
from n = 1 to higher levels. Hence, some of the incident wavelengths will be absent in A.
Photons emerging due to transitions of electrons to their ground state contains visible region wavelengths if transition
is done into n = 2. If transition is done into n > 2, photons contain I.R wavelengths.
Therefore, A, C, D is the correct choice.

36. (A, B)

An = πr 2 = π(r0 n 2 ) 2 = πr0 2 n 4
An
ln = ln(n 4 ) = 4ln n
A1

37. (A, C)
If K < 20.4 ev
∆E = {0,10.2 ev, 12.09 ev}
∆E = {0, 7 ev}
loss = 0
so elastic collision
if (K.E.) > 20.4ev
then if loss = 0 then elastic & otherwise inelastic collision
38. (A, D)
As the potential difference is increased, the kinetic energy is increased. The total energy of X-rays emitted will also
increase hence intensity will increase. Also, the shorter wavelengths will also decrease.
39. (A, B, C)
(A) Minimum wavelength will correspond to maximum energy i.e., from ∞ to ground state
ΔE = 19.9 KeV
 1.24 × 104
∴ λ min = Å0.62 Å = 62 pm.
19.9 × 103
(B) Energy of the characteristic X-rays will be less than corresponding to ∞ to k-shell, hence less than 19.9 KeV.
(C) Energy required to remove as electron from L shell is 19.9 keV so Lα x-ray may be emitted.

40. (B)

41. (C, D)
(i) Maximum energy from one atom
4800 1 48
Rch × = Rch
49 100 49
 1 
(ii) ∆E = 13.6 1 − 2

eV
 ( n + 1) 
48  1 
× 13.6 = 13.6 × 1 −
49  (n + 1) 2  ⇒ n = 6.
 
(iii) Since n = 6 is the excited state. An atom can emit a maximum transition of 6.
n=6
∴ Total no. of photons = 6 × 100 = 600.
42. (A, B, C, D)
n(n − 1)
Number of lines possible =
2
1 1
En2 − En1 = E0 Z 2  2 − 2 
 n1 n2 
So, E4 – E3 = 0.66 eV
E0 = 13.6 eV = Ionization energy of hydrogenation.
n(n − 1)
2
43. (A, B)
Fact based
44. (A, B, C, D)
Fact based

45. (A, B)
For the third line of Balmer series n1 = 2, n2 = 5
1  1 1   1 1  21RZ 2
= RZ 2  2 − 2  = RZ 2  2 − 2  = .........(i )
λ n  2 5  100
 1 n2 
1 1
= ...........(ii )
λ 108.5 × 10−9
From (i) and (ii) (Z = 2)
Binding energy of an electron in the ground state of
13.6 Z 2
hydrogen = = 54.4eV(n = 1)
n2
46. (A, B, C, D)
Under normal conditions total energy, potential energy and kinetic energy in ground state and first excited state are
–13.6 eV, –27.2 eV, 13.6 eV, –3.4 eV, – 6.8 eV and 3.4 eV respectively. If potential energy in ground state is taken
to be zero, then kinetic energy will remain unchanged but potential and total energies are increased by
27.2 eV. Therefore, the new values are 13.6 eV, 0, 13.6 eV, 23.8 eV, 20.4 eV and 3.4 eV respectively.
47. (B, C)
1
U∝ (with negative sign)
n2
1
K∝ and L ∝ n
n
Therefore, as n is increased U (PE) and L (angular momentum) increase while K (KE) and v (linear speed) decrease.
Further, linear speed is decreasing and radius is increasing. Therefore, angular speed (ω = v/r) will also decrease.

48. (A, B, C, D)
1 1
E∝ 2
,P ∝ r ∝ n2
n n
1 P
EPr ∝ , ∝ n, Er ∝ n0 and Pr ∝ n
n E

49. (A, D)
Let the spectral line in the transition from n = p to n = q in hydrogen atom is same as the transition from n = a to
n = b in helium atom. Then
 1 1   1 1 
 2 − 2  = 4  2 − 2 
q p  b a 
2 2
1 1
2  2
2
− 2 =  − 
q p b a
This will be satisfied if b = 2q and a = 2p
i.e, a and b should be even integers and greater than 1.

50. (A, B)
− Ke 2 1
U= or U = −
3r 3
r3
dU 1
F =− or F ∝ −
dr r4
Negative sign indicates that direction of force is radially inwards which provides the necessary centripetal force.
A mv 2
∴ = …. (1)
r4 r
nh
and mvr = …. (2)
2π

Here A is a constant solving these two equations we get 1/2 mV2 or kinetic energy ∝ n6 / m3 .

51. (26.00)
1 1 1
Kα = = R ( Z − 1) 2  2 − 2 
λα 1 2 
 4  hc 
λα = 2 λ = 
3R ( Z − 1)  eV 
hc hc λ V
λ1 = , λ2 = ⇒ 1 = 2 ⇒ λ1 = 2λ 2
eV1 eV2 λ 2 V1
13
( K α − λ1 ) = ( Kα − λ 2 )
10
 13   13 1   λ1 
K α  − 1 = λ1  −  λ 2 = 2 
 10   10 2   
 3 8
K α   = λ1  
 10   10 
4 hc
2
×3 = × 8 ⇒ eV1 = hcR ( Z − 1) 2
3R( Z − 1) eV1
So, Z = 26

52. (2.00)
Since we obtain 6 emission lines, it means electron comes from 4th orbit energy emitted is equal to less than and
more than 2.7eV. So it can be like this.
E4 – E2 = 2.7eV,
E4 – E3 < 2.7eV,
E4 – E1 > 2.7eV
So, nB = 2 (nearest integer)

53. (3.80)
54. (0.00)
1 υ
f= =
T 2πr
υ  nh 
frl = ×r×l l = 
2πr  2π 
υ nh
= ×
2π 2π
z
v0 ×
= n × nh = independent of ‘n’ is
2π 2π
So, frl α n0

55. (3.00)
First excitation potential of
He+ = 10.2 × z2
z = 2 for He+
n = 40.8 V
Ionization potential for
Li++ = 13.6 × z2
m = 122.4
m 122.4
= =3
n 40.8

56. (4.00)
 1
Energy of neutron is mV 2
2
1
The condition for inelastic collision is ⇒ mV 2 > 2∆E
2
1
⇒ ∆E = mV 2
4
∆E is the energy absorbed
Energy required for first excited state is 10.2 eV
∴ ∆E < 10.2 eV
1 4 × 10.1
10.2 < mV 2 ⇒ Vmin = eV
4 m
V = 6 × 104 m/sec.

57. (2.00)
 1 1 
∆E1 = −13.6 z 2  2 − 2  = 2.75 eV
5 4 
 1 1 
∆E2 = −13.6 z 2  2 − 2  = 5.95 eV
4 3 
We know
ev0 = hf − φ
Given that V0 z = 3.95 V
eV0 = ∆E2 − φ
φ = 5.95 − 3.95 = 2eV

58. (12.00)
In energy units, 1 rydberg = 13.6 eV. The energy needed to detach the electron is 4 × 13.6 eV. The energy in the
ground state is, therefore, E1 = – 4 × 13.6 eV. The energy of the first excited state (n = 2) is
E
E2 = = 13.6 eV
4
E2 = 40.84 eV. The wavelength of the radiation emitted is
hc
λ= = 303.92 Å
∆E
The energy of a hydrogen-like ion in ground state is
E = Z 2E0 where Z = atomic number and E0 = –13.6 eV.
a
Thus, Z = 2. The radius of the first orbit is 0 where
Z
a0 = 5 × 10–11 m. Thus,
a
r = 0 × 2.5 × 10–11 m
Z
N 303.92 × 10−10
So = −10
= 1215.68 = 12.15 × 10−2
r 0.25 × 10

59. (4.00)
 1 
Ephoton = 13.6 1 −  eV = 13.0eV
 25 
E/c = mv (Momentum conserved)
 E (13)(1.6 × 10−19 )
v= = = 4 m/sec
mc (1.67)(10−27 )(3)(108 )
60. (6.00)
When electron jumps from nth state to ground state, number of possible emission lines
n(n − 1)
=
2
Here, number of possible emission lines
(n − 1)(n − 2)
= = 10 (given)
2
On solving n = 6

61. (8.00)
A = v2/r
Z2
a∝
(1/ Z )
V ∝Z
1
r∝
Z
Thus, a ∝ Z3
3
a1  Z1   2 
= =  = 8
a2  Z 2   1 

62. (2.00)
The shortest wavelength of Brackett series is corresponding to transition of election between
n1 = 4 and n2 = ∞
Similarly, the shortest wavelength of Balmer series is corresponding to transition of electron between
n1 = 2 and n2 = ∞
 13.6   13.6 
Thus, we have (Z2)  = 
 16   4 
Or Z = 2

63. (8.00)
V2 (Z 2 )
a= Or a ∝ ∴a ∝ Z 3
r (1/ Z )
For singly ionized helium atom, Z = 2
And doubly ionized lithium atom, Z = 3

64. (5.00)
µ i µ (qf ) µ0 (q )(v / 2πr )
B= 0 = 0 =
2r 2r r
1
v 1
B ∝ 2 ∝ n4 ∴ B∝
r n n5

65. (4.00)
 1 1 
4.896 = 13.6Z 2  2 − 
 n (n + 1) 2 

Z2
13.6 = 13.6
n2
4.896 n2 n 4
=1− 2
= =
13.6 (n + 1) n +1 5

66. (255.00)
Let mass of hydrogen atom be m and initial velocity of helium be u.
Therefore from momentum conservation (for maximum loss of energy)
4u
4mu = 4mv1 + mv1 ⇒ v1 =
5
Loss in kinetic energy
2
1 1  4u   1  4  1
= 4mu 2 − 5m   =  4mu 2 1 −  = ( KEi )
2 2  5  2  5  5
Loss in KE must be equal to the sum of excitation energy of helium and hydrogen atom
1
10.2eV + 40.8eV = KEi
5
KEi = (51) × 5eV = 255eV

67. (2.00)
|TE| = (BE) PE = 2[TE]
−13.6 × 12
TE = (eV) = –3.4eV
22
PE = 2×(TE) = 2×(TE) = – 6.8eV
KE = |TE| = 3.4 eV

68. (79.4)
When one electron is removed the remaining atom is hydrogen like atom with Z = 2
Binding energy of such atom will be
= 13.6 × 22 = 54.4 eV
Total energy needed = 54.4 + 25 = 79.4 eV

69. (9.00)
From the given conditions:
En – E2 = (10.2 + 17)eV = 27.2eV ...(1)
and En – E3 = (4.25 + 5.95) eV = 10.2 eV ...(2)
Equation (1) – (2) gives
1 1
E3 – E2 = 17.0 eV or Z2(13.6)  −  = 17.0
4 9
⇒ Z2(13.6) (5/36) = 17.0 ⇒ Z2 = 9 ⇒ Z = 3
1 1 
From equation (1) Z2(13.6)  − 2  = 27.2
4 n 
 1 1  1 1
⇒ (3)2 (13.6)  − 2  = 27.2 ⇒ − 2 = 0.222
4 n  4 n
⇒ 1/n2 = 0.0278 ⇒ n2 = 36 ⇒ n = 6
So n + Z = 9

70. (7.00)
1  1 1 1 R
= R 2 −  ⇒ =
λB  (2) ∞  λB 4

1  1 1  7R 7
= R −  = ;λB : λP =
λP  9 16  144 36

71. (37.00)
hc 12400
∴ λmin = = Å
eV V
12400
At 40 kV : λmin = = 0.31Å
40000
Wavelength of Kα is independent of applied potential.
3 hc
For Kα X–ray : (13.6)(Z – 1)2 = E =
4 λkα
1216
λ kα = Å and given that
( Z − 1) 2
1216
λ kα = 3 λ min ⇒ = 3 0.31
( Z − 1) 2
1216
⇒ (Z – 1)2 =  1308 ⇒ Z – 1 = 36 ⇒ Z = 37
0.93

72. (30.6)
13.6( Z ) 2
Energy required to remove an electron from an orbit is + eV.
n2
So, to remove the electron from the first excited state of Li2+ is
+13.6 × 32
E= = 3.4 9 = +30.6 eV
22
73. (0.25)
 m2  m2
(rm) =   (0.53 Å) = (n 0.53)Å ∴ =n
 z  z
 
m = 5 for 100Fm257 (the outermost shell) and z =100
(5)2 1
∴n= =
100 4

74. (6.00)
 2 2
1 λ  Z −1 1  Z −1
∝ ( Z − 1)2 ∴ 1 =  2  ⇒ =  2 
λ λ 2  Z1 − 1  4  11 − 1 
Solving this, we get Z2 = 6

75. (B)
I-(P, R); II-(Q, S); III-(Q, S); IV-(P, R)
1 1
E∝ 2
,V ∝ and r ∝ n 2
n n
11 1
(a) Epr ∝ × × n 2 or Epr ∝
2
n n n
p 1 2 p
(b) ∝ × n or ∝ n
E n E
1 2 p
(c) Er ∝ × n or ∝ n
n E
1 2
(d) Pr ∝ × n or pr ∝ n
n

76. (A)
I → S; II → R; III → P; IV → Q

77. (B)
78. (C)
I-P, Q; II-T; C-T
1 1 v
v ∝ , KE ∝ 2 , J ∝ n, ω =
n n r
1 1
But r ∝ n 2 and v ∝ ∴ω∝ 3
n n

79. (B)
I-S; II-R; III-Q; IV-P
For given atomic number, energy and hence frequency of K–series is more than L–series. In one series also β–line
has more energy or frequency compared to that of α–line.

80. (D)
For Balmer series, n1 = 2, n2 = 3, 4, ........
(lower) (higher)
∴ In transition (VI), Photon of Balmer series is absorbed.

81. (C)
In transition II
E2 = –3.4 eV, E4 = –0.85 eV
ΔE = 2.55 eV
hc hc
∆E = ⇒λ=
λ ∆E
λ = 487 nm.
82. (D)
Wavelength of radiation = 103 nm = 1030 Å
 12400
∆E =
o
1030 A
 12.0eV
So, difference of energy should be 12.0 eV (approx.)
Hence n1 = 1 and n2 = 3
(–13.6) eV (–1.51) eV ≈ 12.0 eV
∴ Transition is V.

83. (A)
nh
mvr = … (i)
2n
mv 2
= qVB … (ii)
r
From (i) and (ii)
nh
r=
2πeB

84. (C)
nh
∵ r=
2πeB
nh nheB
So, v = = ,
2πmr 2πm 2
for minimum n = 1.

85. (B)
Ground state energy (in eV) is Ei.
Given condition, E2, – Ei = 204 eV
 1 
E1  2 − 1 = 204eV …. (1)
 4n 
and E2n – 2n = 40.8eV
 1 1 3E
E1  2 − 2  = − 21 = 40.8 eV …. (2)
 4n n  4n
By dividing equation (1) and (2), we get.
 1 
1 − 4n 2 
  =5
 3 
 4n 2 
 
So, solving we get, n = 2

86. (C)
4
E1 = − n 2 (40.8)eV = −217.6eV (put n = 2)
3

87. (B)
E1 = −(13.6) Z2 = −217.6eV
 E E1
Emin = E2 n − E2 n −1 = 2
− (Put n = 2)
4n (2n − 1) 2
E1 E1 −7 E1
= − = = +10.58 eV
16 9 144

88. (B)

89. (C)
E0
Let us assume that ions were in nth state before radiation of photon of energy 144 As 10 distinct lines are observed
25
in to emission so ions were excited to 5th state after photon absorption
144 E0
Now, E5 − En =
225
1 1  144 E0
Z 02 E0  2 − 2  =
5 n  225
1 1
= E0  2 − 2  × 32
3 5 
Comparing Z10 = 3 and n = 3
Answer will be 2nd excited state before radiation and 4th excited state after radiation.

90. (D)

91. (A)
The energy of electron in the nth state of He+ ion of atomic number Z is given by
Z2
En = −(13.6eV )
n2
For He+ ion, Z = 2, Therefore
(13.6eV ) × (2) 2
En = −
n2
54.4
En = − eV …. (i)
n2
The energies E1 & E2 of the two emitted photons in eV are
12431
E1 = eV = 11.4eV
1085
12431
E2 = eV = 40.9eV
304
Thus, total energy
E = E1 + E2 = 11.4 + 40.9 = 52.3eV
Let n be the principal quantum number of the excited state. Using equation (i), we have for the transition from n = n
to n =1
But E = 52.3eV. Therefore
 1 
52.3eV = 54.4eV × 1 − 2  ⇒ n = 5 ;
 n 
Also, the energy of the incident electron = 100 eV (given) the energy supplied to H e+ ion
1 52.3
⇒ 1− 2
= = 0.96
n 54.4
 Which gives, n2 = 25 or n = 5
The energy of the incident electron = 100 eV (given). The energy supplied to He+ ion = 52.3 eV.
Therefore, the energy of the electron left after the collision =100 – 52.3 = 47.7eV

92. (D)
As emission spectrum of the atom shows 10 different lines, so the highest excited state of the electron is n = 5.

93. (B)
64 64
As some lines have energy less then E0 and some have more than E0 and some have equal to it. So initial
225 225
state of electron can be n = 2 or n = 3.
Again, if initial level is n = 2, then
 E   E  64
Z2 − 0  − Z2 − 0  = E0 , is in fraction (not possible)
 25   4  225

94. (A)
 E   E  64
If n = 3, then Z 2  − 0  − Z 2  − 0  = E0 , Z = 2
 25   9  225

95. (A)
1 o
From figure = = 8.06( A) −1
λ min
If the bombarding electrons have KE of E (in eV units) then
12420
= E [Where λmin is in Å]
λ min
∴ E = 12420 × 8.06  105 eV
∴ Accelerating potential difference is 105 volt.

1
96. (B) ka will have larger wavelength (and hence smaller )
λ
1 o o
∴ = 4.34( A) −1 ⇒ λ k α = 0.23 A
λkα

97. (C)
We know
1 1 1 
∴ = ( Z − 1) 2  2 − 2 
λkα 1 2 
3
⇒ 4.34 × 1010 m = ( Z − 1) 2 × 1.09 × 107 ×
4
∴ (Z – 1) = 73
Z = 74