BOHR’S THEORY AND PHYSICS OF ATOM

CHAPTER 43
0 h2 A 2 T 2 (ML2 T 1 )2 M2L4 T 2
1. a0 =  2
 L
me 2 2
L MLT M(AT) 2
M2L3 T 2
a0 has dimensions of length.
7 2 2
2. We know,   1/  = 1.1  10  (1/n1 – 1/n2 )
a) n1 = 2, n2 = 3
7
or, 1/ = 1.1  10  (1/4 – 1/9)
36 –7
or,  = = 6.54  10 = 654 nm
5  1.1 107
b) n1 = 4, n2 = 5
   1/  = 1.1  107 (1/16 – 1/25)
400 –7
or,  = = 40.404  10 m = 4040.4 nm
1.1 107  9
7
for R = 1.097  10 ,  = 4050 nm
c) n1 = 9, n2 = 10
7
 1/ = 1.1  10 (1/81 – 1/100)
8100 –7
or,  = = 387.5598  10 = 38755.9 nm
19  1.1 107
7
for R = 1.097  10 ;  = 38861.9 nm
3. Small wave length is emitted i.e. longest energy
n1 = 1, n2 = 
1  1 
a)  R 2 2 
 n
 1  n 2 

1 1 1 
  1.1 107   
 1  
1 1
 10 7 = 0.909  10 = 90.9  10 = 91 nm.
–7 –8
 = 7

1.1 10 1.1
1  1 
b)  z2R  2 2 
  n1  n2 
1 91 nm
 =  = 23 nm
1.1 10 7 z2 4
1  1 
c)  z2R  2 2 
 n
 1  n 2 

91 nm 91
 =  = 10 nm
z2 9
me 4
4. Rydberg’s constant =
8h3 C02
–31 –19 –34 8 –12
me = 9.1  10 kg, e = 1.6  10 c, h = 6.63  10 J-S, C = 3  10 m/s, 0 = 8.85  10
9.1 1031  (1.6  10 19 )4 7 –1
or, R = = 1.097  10 m
8  (6.63  10 34 )3  3  108  (8.85  10 12 )2
5. n1 = 2, n2 = 
13.6 13.6  1 1 
E = 2
 2
 13.6  2  2 
n1 n2  n1 n2 
= 13.6 (1/ – 1/4) = –13.6/4 = –3.4 eV

43.1
 Bohr’s Theory and Physics of Atom
2 2 2
0h n 0.53n
6. a) n = 1, r =  A
mZe2 Z
0.53  1
= = 0.265 A°
2
13.6z2 13.6  4
=  = –54.4 eV
n2 1
0.53  16
b) n = 4, r = = 4.24 A
2
13.6  4
 = = –3.4 eV
164
0.53  100
c) n = 10, r = = 26.5 A
2
13.6  4
 = = –0.544 A
100
7. As the light emitted lies in ultraviolet range the line lies in hyman series.
1  1 1
 R 2  2 
  n1 n2 
1 7 2 2
 9
= 1.1  10 (1/1 – 1/n2 )
102.5  10
109 102
  1.1 107 (1  1/ n22 )   1.1 107 (1  1/ n22 )
102.5 102.5
1 100 1 1  100
 1 2   2 
n2 102.5  1.1 n2 102.5  1.1
 n2 = 2.97 = 3.
8. a) First excitation potential of
+ 2
He = 10.2  z = 10.2  4 = 40.8 V
++
b) Ionization potential of L1
2
= 13.6 V  z = 13.6  9 = 122.4 V
9. n1 = 4  n2 = 2
n1 = 4  3  2
1  1 1
 1.097  107   
  16 4 
1  1 4  1.097  107  3
  1.097  107  
  16  16
16  10 7 –7
 = = 4.8617  10
3  1.097
–9
= 1.861  10 = 487 nm
n1 = 4 and n2 = 3
1  1 1
 1.097  107   
  16 9 
1  9  16  1.097  107  7
  1.097  107  
  144  144
144
 = = 1875 nm
7  1.097  107
n1 = 3  n2 = 2
1  1 1
 1.097  107   
 9 4

43.2
 Bohr’s Theory and Physics of Atom
7
1 49 1.097  10  5
  1.097  107  
  36  66
36  10 7
 = = 656 nm
5  1.097
10.  = 228 A°
hc 6.63  1034  3  108 –16
E= = = 0.0872  10
 228  10 10
The transition takes place form n = 1 to n = 2
2 –16
Now, ex. 13.6  3/4  z = 0.0872  10
2 0.0872  10 16  4
 z = = 5.3
13.6  3  1.6  10 19
z = 5.3 = 2.3
The ion may be Helium.
q1q2
11. F =
40r 2
[Smallest dist. Between the electron and nucleus in the radius of first Bohrs orbit]
(1.6  10 19 )  (1.6  10 19 )  9  109 –9 –8 –8
= = 82.02  10 = 8.202  10 = 8.2  10 N
(0.53  10 10 )2
12. a) From the energy data we see that the H atom transists from binding energy of 0.85 ev to exitation
energy of 10.2 ev = Binding Energy of –3.4 ev.
–0.85 eV
So, n = 4 to n = 2
7 –1.5 eV
b) We know = 1/ = 1.097  10 (1/4 – 1/16)
–3.4 eV
16 –7
 = = 4.8617  10 = 487 nm. –13.6 eV
1.097  3  107
13. The second wavelength is from Balmer to hyman i.e. from n = 2 to n = 1
n1 = 2 to n2 = 1
1  1 1
 R 2  2 
 n
 1 n 2 

1  1 1 1 
  1.097  107  2  2   1.097  107   1
 2 1  4 
4
   10 7
1.097  3
–7 –9
= 1.215  10 = 121.5  10 = 122 nm.
13.6
14. Energy at n = 6, E = = –0.3777777
36
Energy in groundstate = –13.6 eV
Energy emitted in Second transition = –13.6 –(0.37777 + 1.13)
= –12.09 = 12.1 eV
b) Energy in the intermediate state = 1.13 ev + 0.0377777
13.6  z2 13.6
= 1.507777 = 2

n n2
13.6
or, n = = 3.03 = 3 = n.
1.507
15. The potential energy of a hydrogen atom is zero in ground state.
An electron is board to the nucleus with energy 13.6 ev.,
Show we have to give energy of 13.6 ev. To cancel that energy.
Then additional 10.2 ev. is required to attain first excited state.
Total energy of an atom in the first excited state is = 13.6 ev. + 10.2 ev. = 23.8 ev.
43.3
 Bohr’s Theory and Physics of Atom
nd
16. Energy in ground state is the energy acquired in the transition of 2 excited state to ground state.
nd
As 2 excited state is taken as zero level.
hc 4.14  10 15  3  108 1242
E=   = 27 ev.
1 46  10 9 46
Again energy in the first excited state
hc 4.14  10 15  3  108
E=  = 12 ev.
II 103.5
17. a) The gas emits 6 wavelengths, let it be in nth excited state.
n(n  1) th
 = 6  n = 4  The gas is in 4 excited state.
2
n(n  1)
b) Total no.of wavelengths in the transition is 6. We have = 6  n = 4.
2
nh 2 nh hn
18. a) We know, m  r =  mr w = w=
2 2 2  m  r 2
1 6.63  10 34 17 17
= = 0.413  10 rad/s = 4.13  10 rad/s.
2  3.14  9.1 10 31  (0.53)2  10 20
19. The range of Balmer series is 656.3 nm to 365 nm. It can resolve  and  +  if / = 8000.
656.3  365
 No.of wavelengths in the range = = 36
8000
Total no.of lines 36 + 2 = 38 [extra two is for first and last wavelength]
20. a) n1 = 1, n2 = 3, E = 13.6 (1/1 – 1/9) = 13.6  8/9 = hc/
13.6  8 4.14  1015  3  108 4.14  3  10 7 –7
or,   = 1.027  10 = 103 nm.
9  13.6  8
b) As ‘n’ changes by 2, we may consider n = 2 to n = 4
1242
then E = 13.6  (1/4 – 1/16) = 2.55 ev and 2.55 = or  = 487 nm.

V0
21. Frequency of the revolution in the ground state is
2r0
[r0 = radius of ground state, V0 = velocity in the ground state]
V0
Frequency of radiation emitted is =f
2r0
C2r0
 C = f   = C/f =
V0
C2r0
= = 45.686 nm = 45.7 nm.
V0
–5
22. KE = 3/2 KT = 1.5 KT, K = 8.62  10 eV/k, Binding Energy = –13.6 (1/ – 1/1) = 13.6 eV.
According to the question, 1.5 KT = 13.6
–5
 1.5  8.62  10  T = 13.6
13.6 5
 T= = 1.05  10 K
1.5  8.62  10 5
+
No, because the molecule exists an H2 which is impossible.
–5
23. K = 8.62  10 eV/k
K.E. of H2 molecules = 3/2 KT
Energy released, when atom goes from ground state to no = 3
 13.6 (1/1 – 1/9)  3/2 KT = 13.6(1/1 – 1/9)
–5 13.6  8
 3/2  8.62  10 T =
9
5 4 4
 T = 0.9349  10 = 9.349  10 = 9.4  10 K.
43.4
 Bohr’s Theory and Physics of Atom
–8
24. n = 2, T = 10 s
me 4
Frequency =
402n3h3
4o2n3h3 4  (8.85)2  23  (6.63)3 10 24  10 102
So, time period = 1/f =  
me 4
9.1 (1.6) 4
10 76
–19
= 12247.735  10 sec.
10 8 5
No.of revolutions = = 8.16  10
12247.735  10 19
6
= 8.2  10 revolution.
25. Dipole moment ()
= n i A = 1  q/t A = qfA
me 4 me5  ( r02n2 )
= e  ( r02n2 ) 
402h3n3 402h3n3
(9.1 10 31 )(1.6  10 19 )5    (0.53)2  10 20  1
=
4  (8.85  10 12 )2 (6.64  10 34 )3 (1)3
–20 –24 2
= 0.0009176  10 = 9.176  10 A-m.
e  me 4  rn2n2
26. Magnetic Dipole moment = n i A =
402h3n3
nh
Angular momentum = mvr =
2
Since the ratio of magnetic dipole moment and angular momentum is independent of Z.
Hence it is an universal constant.
e5  m  r02n2 2 (1.6  10 19 )5  (9.1 10 31 )  (3.14)2  (0.53  10 10 )2
Ratio =  
240h3n3 nh 2  (8.85  1012 )2  (6.63  10 34 )4  12
10
= 8.73  10 C/kg.
1242
27. The energies associated with 450 nm radiation = = 2.76 ev
450
1242
Energy associated with 550 nm radiation = = 2.258 = 2.26 ev.
550
The light comes under visible range
Thus, n1 = 2, n2 = 3, 4, 5, ……
2 2
E2 – E3 = 13.6 (1/2 – 1/3 ) = 1.9 ev
E2 – E4 = 13.6 (1/4 – 1/16) = 2.55 ev
E2 – E5 = 13.6 (1/4 – 1/25) = 2.856 ev
Only E2 – E4 comes in the range of energy provided. So the wavelength corresponding to that energy
will be absorbed.
1242
= = 487.05 nm = 487 nm
2.55
487 nm wavelength will be absorbed.
28. From transitions n =2 to n =1.
E = 13.6 (1/1 – 1/4) = 13.6  3/4 = 10.2 eV
Let in check the transitions possible on He. n = 1 to 2
E1 = 4  13.6 (1 – 1/4) = 40.8 eV [E1 > E hence it is not possible]
n = 1 to n = 3
E2 = 4  13.6 (1 – 1/9) = 48.3 eV [E2 > E hence impossible]
Similarly n = 1 to n = 4 is also not possible.
n = 2 to n = 3
E3 = 4  13.6 (1/4 – 1/9) = 7.56 eV

43.5
 Bohr’s Theory and Physics of Atom
n = 2 to n = 4
E4 = 4  13.6 (1/4 – 1/16) = 10.2 eV
As, E3 < E and E4 = E
Hence E3 and E4 can be possible.
29.  = 50 nm
Work function = Energy required to remove the electron from n1 = 1 to n2 = .
E = 13.6 (1/1 – 1/) = 13.6
hc
 13.6 = KE

1242
  13.6 = KE  KE = 24.84 – 13.6 = 11.24 eV.
50
30.  = 100 nm
hc 1242
E=  = 12.42 eV
 100
a) The possible transitions may be E1 to E2
E1 to E2, energy absorbed = 10.2 eV
Energy left = 12.42 – 10.2 = 2.22 eV
hc 1242
2.22 eV =  or  = 559.45 = 560 nm
 
E1 to E3, Energy absorbed = 12.1 eV
Energy left = 12.42 – 12.1 = 0.32 eV
hc 1242 1242
0.32 =  or = = 3881.2 = 3881 nm
  0.32
E3 to E4, Energy absorbed = 0.65
Energy left = 12.42 – 0.65 = 11.77 eV
hc 1242 1242
11.77 =  or = = 105.52
  11.77
b) The energy absorbed by the H atom is now radiated perpendicular to the incident beam.
hc 1242
 10.2 = or  = = 121.76 nm
 10.2
hc 1242
 12.1 = or  = = 102.64 nm
 12.1
hc 1242
 0.65 = or  = = 1910.76 nm
 0.65
31. = 1.9 eV
a) The hydrogen is ionized
n1 = 1, n2 = 
2 2
Energy required for ionization = 13.6 (1/n1 – 1/n2 ) = 13.6
hc
 1.9 = 13.6   = 80.1 nm = 80 nm.

b) For the electron to be excited from n1 = 1 to n2 = 2
2 2 13.6  3
 E = 13.6 (1/n1 – 1/n2 ) = 13.6(1 – ¼) =
4
hc 13.6  3
 1.9    = 1242 / 12.1 = 102.64 = 102 nm.
 4
32. The given wavelength in Balmer series.
The first line, which requires minimum energy is from n1 = 3 to n2 = 2.
 The energy should be equal to the energy required for transition from ground state to n = 3.
i.e. E = 13.6 [1 – (1/9)] = 12.09 eV
 Minimum value of electric field = 12.09 v/m = 12.1 v/m

43.6
 Bohr’s Theory and Physics of Atom
33. In one dimensional elastic collision of two bodies of equal masses.
The initial velocities of bodies are interchanged after collision.
 Velocity of the neutron after collision is zero.
Hence, it has zero energy.
34. The hydrogen atoms after collision move with speeds v1 and v2.
mv = mv1 + mv2 …(1)
1 1 1
mv 2  mv12  mv 22  E …(2)
2 2 2
2 2
From (1) v = (v1 + v2) = v12  v 22  2v1v 2
2
From (2) v = v12  v 22  2E / m
2E
= 2v1v 2  …(3)
m
(v1  v 2 )2  (v1  v 2 )2  4v1v 2
2
 (v1 – v2) = v – 4E/m
For minimum value of ‘v’
2
v1 = v2  v – (4E/m) = 0
2 4E 4  13.6  1.6  10 19
 v = 
m 1.67  10 27
4  13.6  1.6  10 19 4
 v= 27
= 7.2  10 m/s.
1.67  10
2
35. Energy of the neutron is ½ mv .
2
The condition for inelastic collision is  ½ mv > 2E
2
 E = ¼ mv
E is the energy absorbed.
Energy required for first excited state is 10.2 ev.
 E < 10.2 ev
2 4  10.2
 10.2 ev < ¼ mv  Vmin = ev
m
10.2  1.6  1019  4 4
v= = 6  10 m/sec.
1.67  10 27
36. a)  = 656.3 nm
hc 1 h 6.63  10 34 –25 –27
Momentum P = E/C =   = = 0.01  10 = 1  10 kg-m/s
 c  656.3  10 9
–27 –27
b) 1  10 = 1.67  10  v
 v = 1/1.67 = 0.598 = 0.6 m/s
–27 2 0.3006  10 27 –9
c) KE of atom = ½  1.67  10  (0.6) = 19
ev = 1.9  10 ev.
1.6  10
37. Difference in energy in the transition from n = 3 to n = 2 is 1.89 ev.
Let recoil energy be E.
2 2 –19
½ me [V2 – V3 ] + E = 1.89 ev  1.89  1.6  10 J
1  2187 2  2187 2 
 9.1 1031 
–19
      E = 3.024  10 J
2   2   3  
–19 –25
 E = 3.024  10 – 3.0225  10
38. n1 = 2, n2 = 3
Energy possessed by H light
= 13.6 (1/n12 – 1/n22) = 13.6  (1/4 – 1/9) = 1.89 eV.
For H light to be able to emit photoelectrons from a metal the work function must be greater than or
equal to 1.89 ev.
43.7
 Bohr’s Theory and Physics of Atom
39. The maximum energy liberated by the Balmer Series is n1 = 2, n2 = 
2 2
E = 13.6(1/n1 – 1/n2 ) = 13.6  1/4 = 3.4 eV
3.4 ev is the maximum work function of the metal.
40. Wocs = 1.9 eV
The radiations coming from the hydrogen discharge tube consist of photons of
energy = 13.6 eV.
+ –
Maximum KE of photoelectrons emitted
= Energy of Photons – Work function of metal.
= 13.6 eV – 1.9 eV = 11.7 eV
41.  = 440 nm, e = Charge of an electron,  = 2 eV, V0 = stopping potential.
hc 4.14  10 15  3  108
We have,    eV0   2eV  eV0
 440  10 9
 eV0 = 0.823 eV  V0 = 0.823 volts.
42. Mass of Earth = Me = 6.0  1024 kg
30
Mass of Sun = Ms = 2.0  10 kg
11
Earth – Sun dist = 1.5  10 m
nh 2 2 2 n2h2
mvr = or, m v r = …(1)
2 42

GMeMs Mev 2 2
 or v = GMs/r …(2)
r2 r
Dividing (1) and (2)

2 n2h2
We get Me r =
42GMs
for n = 1

h2 –138 –138
r= = 2.29  10 m = 2.3  10 m.
4 GMsMe2
2

Me2  r  4  2  G  Ms
b) n2 = = 2.5  10 .
74

h2
nh
43. meVr = …(1)
z
GMnMe me V 2 GMn
2
   2 …(2)
r r r
Squaring (2) and dividing it with (1)
m2e v 2r 2 n2h2r 2 n2h2r n2h2r
2
 2
 me r = 2
r=
 4 Gmn 4 Gmn 4 Gmnme2
2

nh
= from (1)
2rme
nh4 2 GMnM2e 2GMnMe
= 2 2
=
2Me n h nh
1 1 (2GMnMe )2 42 G2Mn2M3e
KE = me V 2  me 
2 2 nh 2n2h2
GMnMe GMnMe 42GMnM2e 42G2Mn2M3e
PE =  
r n2h2 n2h2
22 G2Mn2M3e
Total energy = KE + PE 
2n2h2
43.8
 Bohr’s Theory and Physics of Atom
44. According to Bohr’s quantization rule
nh
mvr =
2
‘r’ is less when ‘n’ has least value i.e. 1
nh
or, mv = …(1)
2R
mv
Again, r = , or, mv = rqB …(2)
qB
From (1) and (2)
nh
rqB = [q = e]
2r
2 nh
r =  r = h / 2 eB [here n = 1]
2eB
nh
b) For the radius of nth orbit, r = .
2eB
nh mv
c) mvr = ,r=
2 qB
Substituting the value of ‘r’ in (1)
mv nh
mv  
qB 2
nheB
 m2 v 2  [n = 1, q = e]
2
heB heB
 v2   or v = .
2m2 2m2
45. even quantum numbers are allowed
n1 = 2, n2 = 4  For minimum energy or for longest possible wavelength.
 1 1  1 1 
E = 13.6  2  2   13.6  2  2  = 2.55
n
 1 n 2   2 4 
hc
 2.55 =

hc 1242
=  = 487.05 nm = 487 nm
2.55 2.55
46. Velocity of hydrogen atom in state ‘n’ = u
Also the velocity of photon = u
But u << C
Here the photon is emitted as a wave.
So its velocity is same as that of hydrogen atom i.e. u.
 According to Doppler’s effect
 1 u / c 
frequency v = v 0  
 1 u / c 
u
as u <<< C 1  q
c
 1 u / c   u  u
 v = v0    v 0 1    v = v 0  1  
 1   c  c



43.9