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 Page 1


43.1
BOHR’S THEORY AND PHYSICS OF ATOM
CHAPTER 43
1. a
0
= 
2 2 2 2 1 2 2 4 2
0
2 2 2 2 2 3 2
h A T (ML T ) M L T
L
me L MLT M(AT) M L T
? ?
? ?
?
? ? ?
?
?a
0
has dimensions of length.
2. We know, 1/ ? ? ? = 1.1 ? 10
7
? (1/n
1
2
– 1/n
2
2
)
a) n
1
= 2, n
2
= 3
or, 1/ ? = 1.1 ? 10
7
? (1/4 – 1/9)
or, ? =
7
36
5 1.1 10 ? ?
= 6.54 ? 10
–7
= 654 nm
b) n
1
= 4, n
2
= 5
? 1/ ? ? ? = 1.1 ? 10
7
(1/16 – 1/25)
or, ? = 
7
400
1.1 10 9 ? ?
= 40.404 ? 10
–7
m = 4040.4 nm
for R = 1.097 ? 10
7
, ? = 4050 nm
c) n
1
= 9, n
2
= 10
? 1/ ? = 1.1 ? 10
7
(1/81 – 1/100) 
or, ? =
7
8100
19 1.1 10 ? ?
= 387.5598 ? 10
–7
= 38755.9 nm
for R = 1.097 ? 10
7
; ? = 38861.9 nm ?
3. Small wave length is emitted i.e. longest energy
n
1
= 1, n
2
= ?
a)
2 2
1 2
1 1
R
n n
? ?
?
? ?
?
?
? ?
?
7
1 1 1
1.1 10
1
? ?
? ? ?
? ?
? ? ? ?
? ? = 
7
7
1 1
10
1.1 1.1 10
?
? ?
?
= 0.909 ? 10
–7
= 90.9 ? 10
–8
= 91 nm.
b)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
7 2
1 91 nm
4 1.1 10 z
?
?
?
= 23 nm
c)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
2
91 nm 91
9 z
? = 10 nm ?
4. Rydberg’s constant = 
4
3 2
0
me
8h C ?
m
e
= 9.1 ? 10
–31
kg, e = 1.6 ? 10
–19
c, h = 6.63 ? 10
–34
J-S, C = 3 ? 10
8
m/s, ?
0
= 8.85 ? 10
–12
or, R = 
31 19 4
34 3 8 12 2
9.1 10 (1.6 10 )
8 (6.63 10 ) 3 10 (8.85 10 )
? ?
? ?
? ? ?
? ? ? ? ? ?
= 1.097 ? 10
7
m
–1 ?
5. n
1
= 2, n
2
= ?
E = 
2 2 2 2
1 2 1 2
13.6 13.6 1 1
13.6
n n n n
? ? ? ?
? ? ?
? ?
? ?
= 13.6 (1/ ? – 1/4) = –13.6/4 = –3.4 eV
Page 2


43.1
BOHR’S THEORY AND PHYSICS OF ATOM
CHAPTER 43
1. a
0
= 
2 2 2 2 1 2 2 4 2
0
2 2 2 2 2 3 2
h A T (ML T ) M L T
L
me L MLT M(AT) M L T
? ?
? ?
?
? ? ?
?
?a
0
has dimensions of length.
2. We know, 1/ ? ? ? = 1.1 ? 10
7
? (1/n
1
2
– 1/n
2
2
)
a) n
1
= 2, n
2
= 3
or, 1/ ? = 1.1 ? 10
7
? (1/4 – 1/9)
or, ? =
7
36
5 1.1 10 ? ?
= 6.54 ? 10
–7
= 654 nm
b) n
1
= 4, n
2
= 5
? 1/ ? ? ? = 1.1 ? 10
7
(1/16 – 1/25)
or, ? = 
7
400
1.1 10 9 ? ?
= 40.404 ? 10
–7
m = 4040.4 nm
for R = 1.097 ? 10
7
, ? = 4050 nm
c) n
1
= 9, n
2
= 10
? 1/ ? = 1.1 ? 10
7
(1/81 – 1/100) 
or, ? =
7
8100
19 1.1 10 ? ?
= 387.5598 ? 10
–7
= 38755.9 nm
for R = 1.097 ? 10
7
; ? = 38861.9 nm ?
3. Small wave length is emitted i.e. longest energy
n
1
= 1, n
2
= ?
a)
2 2
1 2
1 1
R
n n
? ?
?
? ?
?
?
? ?
?
7
1 1 1
1.1 10
1
? ?
? ? ?
? ?
? ? ? ?
? ? = 
7
7
1 1
10
1.1 1.1 10
?
? ?
?
= 0.909 ? 10
–7
= 90.9 ? 10
–8
= 91 nm.
b)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
7 2
1 91 nm
4 1.1 10 z
?
?
?
= 23 nm
c)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
2
91 nm 91
9 z
? = 10 nm ?
4. Rydberg’s constant = 
4
3 2
0
me
8h C ?
m
e
= 9.1 ? 10
–31
kg, e = 1.6 ? 10
–19
c, h = 6.63 ? 10
–34
J-S, C = 3 ? 10
8
m/s, ?
0
= 8.85 ? 10
–12
or, R = 
31 19 4
34 3 8 12 2
9.1 10 (1.6 10 )
8 (6.63 10 ) 3 10 (8.85 10 )
? ?
? ?
? ? ?
? ? ? ? ? ?
= 1.097 ? 10
7
m
–1 ?
5. n
1
= 2, n
2
= ?
E = 
2 2 2 2
1 2 1 2
13.6 13.6 1 1
13.6
n n n n
? ? ? ?
? ? ?
? ?
? ?
= 13.6 (1/ ? – 1/4) = –13.6/4 = –3.4 eV
Bohr’s Theory and Physics of Atom
43.2
6. a) n = 1, r = 
2 2 2
0
2
h n 0.53n
A
Z mZe
?
? ?
?
= 
0.53 1
2
?
= 0.265 A°
? = 
2
2
13.6z 13.6 4
1 n
? ? ?
? = –54.4 eV
b) n = 4, r = 
0.53 16
2
?
= 4.24 A
? ? = 
13.6 4
164
? ?
= –3.4 eV
c) n = 10, r = 
0.53 100
2
?
= 26.5 A
? ? = 
13.6 4
100
? ?
= –0.544 A ?
7. As the light emitted lies in ultraviolet range the line lies in hyman series.
2 2
1 2
1 1 1
R
n n
? ?
? ?
? ?
?
? ?
?
9
1
102.5 10
?
?
= 1.1 ? 10
7
(1/1
2
– 1/n
2
2
)
?
9 2
7 2 7 2
2 2
10 10
1.1 10 (1 1/n ) 1.1 10 (1 1/n )
102.5 102.5
? ? ? ? ? ? ?
?
2 2
2 2
1 100 1 1 100
1
102.5 1.1 102.5 1.1 n n
?
? ? ? ?
? ?
? n
2
= 2.97 = 3.
8. a) First excitation potential of
He
+
= 10.2 ? z
2
= 10.2 ? 4 = 40.8 V
b) Ionization potential of L
1
++
= 13.6 V ? z
2 
= 13.6 ? 9 = 122.4 V
9. n
1
= 4 ? n
2
= 2
n
1
= 4 ? 3 ? 2
7
1 1 1
1.097 10
16 4
? ?
? ? ?
? ?
? ? ?
?
7
7
1 1 4 1.097 10 3
1.097 10
16 16
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
16 10
3 1.097
?
?
?
= 4.8617 ? 10
–7
= 1.861 ? 10
–9
= 487 nm
n
1
= 4 and n
2
= 3
7
1 1 1
1.097 10
16 9
? ?
? ? ?
? ?
? ? ?
?
7
7
1 9 16 1.097 10 7
1.097 10
144 144
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
144
7 1.097 10 ? ?
= 1875 nm
n
1
= 3 ? n
2
= 2
7
1 1 1
1.097 10
9 4
? ?
? ? ?
? ?
? ? ?
Page 3


43.1
BOHR’S THEORY AND PHYSICS OF ATOM
CHAPTER 43
1. a
0
= 
2 2 2 2 1 2 2 4 2
0
2 2 2 2 2 3 2
h A T (ML T ) M L T
L
me L MLT M(AT) M L T
? ?
? ?
?
? ? ?
?
?a
0
has dimensions of length.
2. We know, 1/ ? ? ? = 1.1 ? 10
7
? (1/n
1
2
– 1/n
2
2
)
a) n
1
= 2, n
2
= 3
or, 1/ ? = 1.1 ? 10
7
? (1/4 – 1/9)
or, ? =
7
36
5 1.1 10 ? ?
= 6.54 ? 10
–7
= 654 nm
b) n
1
= 4, n
2
= 5
? 1/ ? ? ? = 1.1 ? 10
7
(1/16 – 1/25)
or, ? = 
7
400
1.1 10 9 ? ?
= 40.404 ? 10
–7
m = 4040.4 nm
for R = 1.097 ? 10
7
, ? = 4050 nm
c) n
1
= 9, n
2
= 10
? 1/ ? = 1.1 ? 10
7
(1/81 – 1/100) 
or, ? =
7
8100
19 1.1 10 ? ?
= 387.5598 ? 10
–7
= 38755.9 nm
for R = 1.097 ? 10
7
; ? = 38861.9 nm ?
3. Small wave length is emitted i.e. longest energy
n
1
= 1, n
2
= ?
a)
2 2
1 2
1 1
R
n n
? ?
?
? ?
?
?
? ?
?
7
1 1 1
1.1 10
1
? ?
? ? ?
? ?
? ? ? ?
? ? = 
7
7
1 1
10
1.1 1.1 10
?
? ?
?
= 0.909 ? 10
–7
= 90.9 ? 10
–8
= 91 nm.
b)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
7 2
1 91 nm
4 1.1 10 z
?
?
?
= 23 nm
c)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
2
91 nm 91
9 z
? = 10 nm ?
4. Rydberg’s constant = 
4
3 2
0
me
8h C ?
m
e
= 9.1 ? 10
–31
kg, e = 1.6 ? 10
–19
c, h = 6.63 ? 10
–34
J-S, C = 3 ? 10
8
m/s, ?
0
= 8.85 ? 10
–12
or, R = 
31 19 4
34 3 8 12 2
9.1 10 (1.6 10 )
8 (6.63 10 ) 3 10 (8.85 10 )
? ?
? ?
? ? ?
? ? ? ? ? ?
= 1.097 ? 10
7
m
–1 ?
5. n
1
= 2, n
2
= ?
E = 
2 2 2 2
1 2 1 2
13.6 13.6 1 1
13.6
n n n n
? ? ? ?
? ? ?
? ?
? ?
= 13.6 (1/ ? – 1/4) = –13.6/4 = –3.4 eV
Bohr’s Theory and Physics of Atom
43.2
6. a) n = 1, r = 
2 2 2
0
2
h n 0.53n
A
Z mZe
?
? ?
?
= 
0.53 1
2
?
= 0.265 A°
? = 
2
2
13.6z 13.6 4
1 n
? ? ?
? = –54.4 eV
b) n = 4, r = 
0.53 16
2
?
= 4.24 A
? ? = 
13.6 4
164
? ?
= –3.4 eV
c) n = 10, r = 
0.53 100
2
?
= 26.5 A
? ? = 
13.6 4
100
? ?
= –0.544 A ?
7. As the light emitted lies in ultraviolet range the line lies in hyman series.
2 2
1 2
1 1 1
R
n n
? ?
? ?
? ?
?
? ?
?
9
1
102.5 10
?
?
= 1.1 ? 10
7
(1/1
2
– 1/n
2
2
)
?
9 2
7 2 7 2
2 2
10 10
1.1 10 (1 1/n ) 1.1 10 (1 1/n )
102.5 102.5
? ? ? ? ? ? ?
?
2 2
2 2
1 100 1 1 100
1
102.5 1.1 102.5 1.1 n n
?
? ? ? ?
? ?
? n
2
= 2.97 = 3.
8. a) First excitation potential of
He
+
= 10.2 ? z
2
= 10.2 ? 4 = 40.8 V
b) Ionization potential of L
1
++
= 13.6 V ? z
2 
= 13.6 ? 9 = 122.4 V
9. n
1
= 4 ? n
2
= 2
n
1
= 4 ? 3 ? 2
7
1 1 1
1.097 10
16 4
? ?
? ? ?
? ?
? ? ?
?
7
7
1 1 4 1.097 10 3
1.097 10
16 16
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
16 10
3 1.097
?
?
?
= 4.8617 ? 10
–7
= 1.861 ? 10
–9
= 487 nm
n
1
= 4 and n
2
= 3
7
1 1 1
1.097 10
16 9
? ?
? ? ?
? ?
? ? ?
?
7
7
1 9 16 1.097 10 7
1.097 10
144 144
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
144
7 1.097 10 ? ?
= 1875 nm
n
1
= 3 ? n
2
= 2
7
1 1 1
1.097 10
9 4
? ?
? ? ?
? ?
? ? ?
Bohr’s Theory and Physics of Atom
43.3
?
7
7
1 4 9 1.097 10 5
1.097 10
36 66
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
36 10
5 1.097
?
?
?
= 656 nm
10. ? = 228 A°
E = 
hc
?
= 
34 8
10
6.63 10 3 10
228 10
?
?
? ? ?
?
= 0.0872 ? 10
–16
The transition takes place form n = 1 to n = 2
Now, ex. 13.6 ? 3/4 ? z
2
= 0.0872 ? 10
–16
? z
2
= 
16
19
0.0872 10 4
13.6 3 1.6 10
?
?
? ?
? ? ?
= 5.3
z = 5.3 = 2.3
The ion may be Helium.
11. F = 
1 2
2
0
q q
4 r ? ?
[Smallest dist. Between the electron and nucleus in the radius of first Bohrs orbit]
=
19 19 9
10 2
(1.6 10 ) (1.6 10 ) 9 10
(0.53 10 )
? ?
?
? ? ? ? ?
?
= 82.02 ? 10
–9
= 8.202 ? 10
–8
= 8.2 ? 10
–8
N
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.
So, n = 4 to n = 2
b) We know = 1/ ? = 1.097 ? 10
7
(1/4 – 1/16)
? ? = 
7
16
1.097 3 10 ? ?
= 4.8617 ? 10
–7
= 487 nm. ?
13. The second wavelength is from Balmer to hyman i.e. from n = 2 to n = 1
n
1
= 2 to n
2
= 1
2 2
1 2
1 1 1
R
n n
? ?
? ?
? ?
?
? ?
?
7 7
2 2
1 1 1 1
1.097 10 1.097 10 1
4 2 1
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ? ?
?
7
4
10
1.097 3
?
? ? ?
?
= 1.215 ? 10
–7
= 121.5 ? 10
–9
= 122 nm. ?
14. Energy at n = 6, E = 
13.6
36
?
= –0.3777777
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
= 1.507777 = 
2
2 2
13.6 z 13.6
n n
?
?
or, n = 
13.6
1.507
= 3.03 = 3 = n.
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.
–0.85 eV
–1.5 eV
–3.4 eV
–13.6 eV
Page 4


43.1
BOHR’S THEORY AND PHYSICS OF ATOM
CHAPTER 43
1. a
0
= 
2 2 2 2 1 2 2 4 2
0
2 2 2 2 2 3 2
h A T (ML T ) M L T
L
me L MLT M(AT) M L T
? ?
? ?
?
? ? ?
?
?a
0
has dimensions of length.
2. We know, 1/ ? ? ? = 1.1 ? 10
7
? (1/n
1
2
– 1/n
2
2
)
a) n
1
= 2, n
2
= 3
or, 1/ ? = 1.1 ? 10
7
? (1/4 – 1/9)
or, ? =
7
36
5 1.1 10 ? ?
= 6.54 ? 10
–7
= 654 nm
b) n
1
= 4, n
2
= 5
? 1/ ? ? ? = 1.1 ? 10
7
(1/16 – 1/25)
or, ? = 
7
400
1.1 10 9 ? ?
= 40.404 ? 10
–7
m = 4040.4 nm
for R = 1.097 ? 10
7
, ? = 4050 nm
c) n
1
= 9, n
2
= 10
? 1/ ? = 1.1 ? 10
7
(1/81 – 1/100) 
or, ? =
7
8100
19 1.1 10 ? ?
= 387.5598 ? 10
–7
= 38755.9 nm
for R = 1.097 ? 10
7
; ? = 38861.9 nm ?
3. Small wave length is emitted i.e. longest energy
n
1
= 1, n
2
= ?
a)
2 2
1 2
1 1
R
n n
? ?
?
? ?
?
?
? ?
?
7
1 1 1
1.1 10
1
? ?
? ? ?
? ?
? ? ? ?
? ? = 
7
7
1 1
10
1.1 1.1 10
?
? ?
?
= 0.909 ? 10
–7
= 90.9 ? 10
–8
= 91 nm.
b)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
7 2
1 91 nm
4 1.1 10 z
?
?
?
= 23 nm
c)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
2
91 nm 91
9 z
? = 10 nm ?
4. Rydberg’s constant = 
4
3 2
0
me
8h C ?
m
e
= 9.1 ? 10
–31
kg, e = 1.6 ? 10
–19
c, h = 6.63 ? 10
–34
J-S, C = 3 ? 10
8
m/s, ?
0
= 8.85 ? 10
–12
or, R = 
31 19 4
34 3 8 12 2
9.1 10 (1.6 10 )
8 (6.63 10 ) 3 10 (8.85 10 )
? ?
? ?
? ? ?
? ? ? ? ? ?
= 1.097 ? 10
7
m
–1 ?
5. n
1
= 2, n
2
= ?
E = 
2 2 2 2
1 2 1 2
13.6 13.6 1 1
13.6
n n n n
? ? ? ?
? ? ?
? ?
? ?
= 13.6 (1/ ? – 1/4) = –13.6/4 = –3.4 eV
Bohr’s Theory and Physics of Atom
43.2
6. a) n = 1, r = 
2 2 2
0
2
h n 0.53n
A
Z mZe
?
? ?
?
= 
0.53 1
2
?
= 0.265 A°
? = 
2
2
13.6z 13.6 4
1 n
? ? ?
? = –54.4 eV
b) n = 4, r = 
0.53 16
2
?
= 4.24 A
? ? = 
13.6 4
164
? ?
= –3.4 eV
c) n = 10, r = 
0.53 100
2
?
= 26.5 A
? ? = 
13.6 4
100
? ?
= –0.544 A ?
7. As the light emitted lies in ultraviolet range the line lies in hyman series.
2 2
1 2
1 1 1
R
n n
? ?
? ?
? ?
?
? ?
?
9
1
102.5 10
?
?
= 1.1 ? 10
7
(1/1
2
– 1/n
2
2
)
?
9 2
7 2 7 2
2 2
10 10
1.1 10 (1 1/n ) 1.1 10 (1 1/n )
102.5 102.5
? ? ? ? ? ? ?
?
2 2
2 2
1 100 1 1 100
1
102.5 1.1 102.5 1.1 n n
?
? ? ? ?
? ?
? n
2
= 2.97 = 3.
8. a) First excitation potential of
He
+
= 10.2 ? z
2
= 10.2 ? 4 = 40.8 V
b) Ionization potential of L
1
++
= 13.6 V ? z
2 
= 13.6 ? 9 = 122.4 V
9. n
1
= 4 ? n
2
= 2
n
1
= 4 ? 3 ? 2
7
1 1 1
1.097 10
16 4
? ?
? ? ?
? ?
? ? ?
?
7
7
1 1 4 1.097 10 3
1.097 10
16 16
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
16 10
3 1.097
?
?
?
= 4.8617 ? 10
–7
= 1.861 ? 10
–9
= 487 nm
n
1
= 4 and n
2
= 3
7
1 1 1
1.097 10
16 9
? ?
? ? ?
? ?
? ? ?
?
7
7
1 9 16 1.097 10 7
1.097 10
144 144
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
144
7 1.097 10 ? ?
= 1875 nm
n
1
= 3 ? n
2
= 2
7
1 1 1
1.097 10
9 4
? ?
? ? ?
? ?
? ? ?
Bohr’s Theory and Physics of Atom
43.3
?
7
7
1 4 9 1.097 10 5
1.097 10
36 66
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
36 10
5 1.097
?
?
?
= 656 nm
10. ? = 228 A°
E = 
hc
?
= 
34 8
10
6.63 10 3 10
228 10
?
?
? ? ?
?
= 0.0872 ? 10
–16
The transition takes place form n = 1 to n = 2
Now, ex. 13.6 ? 3/4 ? z
2
= 0.0872 ? 10
–16
? z
2
= 
16
19
0.0872 10 4
13.6 3 1.6 10
?
?
? ?
? ? ?
= 5.3
z = 5.3 = 2.3
The ion may be Helium.
11. F = 
1 2
2
0
q q
4 r ? ?
[Smallest dist. Between the electron and nucleus in the radius of first Bohrs orbit]
=
19 19 9
10 2
(1.6 10 ) (1.6 10 ) 9 10
(0.53 10 )
? ?
?
? ? ? ? ?
?
= 82.02 ? 10
–9
= 8.202 ? 10
–8
= 8.2 ? 10
–8
N
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.
So, n = 4 to n = 2
b) We know = 1/ ? = 1.097 ? 10
7
(1/4 – 1/16)
? ? = 
7
16
1.097 3 10 ? ?
= 4.8617 ? 10
–7
= 487 nm. ?
13. The second wavelength is from Balmer to hyman i.e. from n = 2 to n = 1
n
1
= 2 to n
2
= 1
2 2
1 2
1 1 1
R
n n
? ?
? ?
? ?
?
? ?
?
7 7
2 2
1 1 1 1
1.097 10 1.097 10 1
4 2 1
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ? ?
?
7
4
10
1.097 3
?
? ? ?
?
= 1.215 ? 10
–7
= 121.5 ? 10
–9
= 122 nm. ?
14. Energy at n = 6, E = 
13.6
36
?
= –0.3777777
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
= 1.507777 = 
2
2 2
13.6 z 13.6
n n
?
?
or, n = 
13.6
1.507
= 3.03 = 3 = n.
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.
–0.85 eV
–1.5 eV
–3.4 eV
–13.6 eV
Bohr’s Theory and Physics of Atom
43.4
16. Energy in ground state is the energy acquired in the transition of 2
nd
excited state to ground state.
As 2
nd
excited state is taken as zero level.
E = 
15 8
9
1
hc 4.14 10 3 10 1242
46 46 10
?
?
? ? ?
? ?
? ?
= 27 ev.
Again energy in the first excited state
E = 
15 8
II
hc 4.14 10 3 10
103.5
?
? ? ?
?
?
= 12 ev.
17. a) The gas emits 6 wavelengths, let it be in nth excited state.
?
n(n 1)
2
?
= 6 ? n = 4 ? The gas is in 4
th
excited state.
b) Total no.of wavelengths in the transition is 6. We have 
n(n 1)
2
?
= 6 ? n = 4.
18. a) We know, m ? r = 
nh
2 ?
? mr
2
w = 
nh
2 ?
? w = 
2
hn
2 m r ? ? ?
= 
34
31 2 20
1 6.63 10
2 3.14 9.1 10 (0.53) 10
?
? ?
? ?
? ? ? ? ?
= 0.413 ? 10
17
rad/s = 4.13 ? 10
17
rad/s.
19. The range of Balmer series is 656.3 nm to 365 nm. It can resolve ? and ? + ?? if ?/ ?? = 8000.
? No.of wavelengths in the range = 
656.3 365
8000
?
= 36
Total no.of lines 36 + 2 = 38 [extra two is for first and last wavelength]
20. a) n
1
= 1, n
2
= 3, E = 13.6 (1/1 – 1/9) = 13.6 ? 8/9 = hc/ ?
or, 
15 8 7
13.6 8 4.14 10 3 10 4.14 3 10
9 13.6 8
? ?
? ? ? ? ? ?
? ? ? ?
? ?
= 1.027 ? 10
–7
= 103 nm.
b) As ‘n’ changes by 2, we may consider n = 2 to n = 4
then E = 13.6 ? (1/4 – 1/16) = 2.55 ev and 2.55 = 
1242
?
or ? = 487 nm. ?
21. Frequency of the revolution in the ground state is 
0
0
V
2 r ?
[r
0
= radius of ground state, V
0
= velocity in the ground state]
?Frequency of radiation emitted is 
0
0
V
2 r ?
= f
? C = f ? ? ? = C/f = 
0
0
C2 r
V
?
? ? = 
0
0
C2 r
V
?
= 45.686 nm = 45.7 nm. ?
22. KE = 3/2 KT = 1.5 KT, K = 8.62 ? 10
–5
eV/k, Binding Energy = –13.6 (1/ ? – 1/1) = 13.6 eV.
According to the question, 1.5 KT = 13.6
? 1.5 ? 8.62 ? 10
–5
? T = 13.6
? T = 
5
13.6
1.5 8.62 10
?
? ?
= 1.05 ? 10
5
K
No, because the molecule exists an H
2
+
which is impossible.
23. K = 8.62 ? 10
–5
eV/k
K.E. of H
2
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)
? 3/2 ? 8.62 ? 10
–5
T = 
13.6 8
9
?
? T = 0.9349 ? 10
5
= 9.349 ? 10
4
= 9.4 ? 10
4
K.
Page 5


43.1
BOHR’S THEORY AND PHYSICS OF ATOM
CHAPTER 43
1. a
0
= 
2 2 2 2 1 2 2 4 2
0
2 2 2 2 2 3 2
h A T (ML T ) M L T
L
me L MLT M(AT) M L T
? ?
? ?
?
? ? ?
?
?a
0
has dimensions of length.
2. We know, 1/ ? ? ? = 1.1 ? 10
7
? (1/n
1
2
– 1/n
2
2
)
a) n
1
= 2, n
2
= 3
or, 1/ ? = 1.1 ? 10
7
? (1/4 – 1/9)
or, ? =
7
36
5 1.1 10 ? ?
= 6.54 ? 10
–7
= 654 nm
b) n
1
= 4, n
2
= 5
? 1/ ? ? ? = 1.1 ? 10
7
(1/16 – 1/25)
or, ? = 
7
400
1.1 10 9 ? ?
= 40.404 ? 10
–7
m = 4040.4 nm
for R = 1.097 ? 10
7
, ? = 4050 nm
c) n
1
= 9, n
2
= 10
? 1/ ? = 1.1 ? 10
7
(1/81 – 1/100) 
or, ? =
7
8100
19 1.1 10 ? ?
= 387.5598 ? 10
–7
= 38755.9 nm
for R = 1.097 ? 10
7
; ? = 38861.9 nm ?
3. Small wave length is emitted i.e. longest energy
n
1
= 1, n
2
= ?
a)
2 2
1 2
1 1
R
n n
? ?
?
? ?
?
?
? ?
?
7
1 1 1
1.1 10
1
? ?
? ? ?
? ?
? ? ? ?
? ? = 
7
7
1 1
10
1.1 1.1 10
?
? ?
?
= 0.909 ? 10
–7
= 90.9 ? 10
–8
= 91 nm.
b)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
7 2
1 91 nm
4 1.1 10 z
?
?
?
= 23 nm
c)
2
2 2
1 2
1 1
z R
n n
? ?
?
? ?
?
?
? ?
? ? = 
2
91 nm 91
9 z
? = 10 nm ?
4. Rydberg’s constant = 
4
3 2
0
me
8h C ?
m
e
= 9.1 ? 10
–31
kg, e = 1.6 ? 10
–19
c, h = 6.63 ? 10
–34
J-S, C = 3 ? 10
8
m/s, ?
0
= 8.85 ? 10
–12
or, R = 
31 19 4
34 3 8 12 2
9.1 10 (1.6 10 )
8 (6.63 10 ) 3 10 (8.85 10 )
? ?
? ?
? ? ?
? ? ? ? ? ?
= 1.097 ? 10
7
m
–1 ?
5. n
1
= 2, n
2
= ?
E = 
2 2 2 2
1 2 1 2
13.6 13.6 1 1
13.6
n n n n
? ? ? ?
? ? ?
? ?
? ?
= 13.6 (1/ ? – 1/4) = –13.6/4 = –3.4 eV
Bohr’s Theory and Physics of Atom
43.2
6. a) n = 1, r = 
2 2 2
0
2
h n 0.53n
A
Z mZe
?
? ?
?
= 
0.53 1
2
?
= 0.265 A°
? = 
2
2
13.6z 13.6 4
1 n
? ? ?
? = –54.4 eV
b) n = 4, r = 
0.53 16
2
?
= 4.24 A
? ? = 
13.6 4
164
? ?
= –3.4 eV
c) n = 10, r = 
0.53 100
2
?
= 26.5 A
? ? = 
13.6 4
100
? ?
= –0.544 A ?
7. As the light emitted lies in ultraviolet range the line lies in hyman series.
2 2
1 2
1 1 1
R
n n
? ?
? ?
? ?
?
? ?
?
9
1
102.5 10
?
?
= 1.1 ? 10
7
(1/1
2
– 1/n
2
2
)
?
9 2
7 2 7 2
2 2
10 10
1.1 10 (1 1/n ) 1.1 10 (1 1/n )
102.5 102.5
? ? ? ? ? ? ?
?
2 2
2 2
1 100 1 1 100
1
102.5 1.1 102.5 1.1 n n
?
? ? ? ?
? ?
? n
2
= 2.97 = 3.
8. a) First excitation potential of
He
+
= 10.2 ? z
2
= 10.2 ? 4 = 40.8 V
b) Ionization potential of L
1
++
= 13.6 V ? z
2 
= 13.6 ? 9 = 122.4 V
9. n
1
= 4 ? n
2
= 2
n
1
= 4 ? 3 ? 2
7
1 1 1
1.097 10
16 4
? ?
? ? ?
? ?
? ? ?
?
7
7
1 1 4 1.097 10 3
1.097 10
16 16
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
16 10
3 1.097
?
?
?
= 4.8617 ? 10
–7
= 1.861 ? 10
–9
= 487 nm
n
1
= 4 and n
2
= 3
7
1 1 1
1.097 10
16 9
? ?
? ? ?
? ?
? ? ?
?
7
7
1 9 16 1.097 10 7
1.097 10
144 144
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
144
7 1.097 10 ? ?
= 1875 nm
n
1
= 3 ? n
2
= 2
7
1 1 1
1.097 10
9 4
? ?
? ? ?
? ?
? ? ?
Bohr’s Theory and Physics of Atom
43.3
?
7
7
1 4 9 1.097 10 5
1.097 10
36 66
? ? ? ? ?
? ? ?
? ?
? ? ?
? ? = 
7
36 10
5 1.097
?
?
?
= 656 nm
10. ? = 228 A°
E = 
hc
?
= 
34 8
10
6.63 10 3 10
228 10
?
?
? ? ?
?
= 0.0872 ? 10
–16
The transition takes place form n = 1 to n = 2
Now, ex. 13.6 ? 3/4 ? z
2
= 0.0872 ? 10
–16
? z
2
= 
16
19
0.0872 10 4
13.6 3 1.6 10
?
?
? ?
? ? ?
= 5.3
z = 5.3 = 2.3
The ion may be Helium.
11. F = 
1 2
2
0
q q
4 r ? ?
[Smallest dist. Between the electron and nucleus in the radius of first Bohrs orbit]
=
19 19 9
10 2
(1.6 10 ) (1.6 10 ) 9 10
(0.53 10 )
? ?
?
? ? ? ? ?
?
= 82.02 ? 10
–9
= 8.202 ? 10
–8
= 8.2 ? 10
–8
N
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.
So, n = 4 to n = 2
b) We know = 1/ ? = 1.097 ? 10
7
(1/4 – 1/16)
? ? = 
7
16
1.097 3 10 ? ?
= 4.8617 ? 10
–7
= 487 nm. ?
13. The second wavelength is from Balmer to hyman i.e. from n = 2 to n = 1
n
1
= 2 to n
2
= 1
2 2
1 2
1 1 1
R
n n
? ?
? ?
? ?
?
? ?
?
7 7
2 2
1 1 1 1
1.097 10 1.097 10 1
4 2 1
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ? ?
?
7
4
10
1.097 3
?
? ? ?
?
= 1.215 ? 10
–7
= 121.5 ? 10
–9
= 122 nm. ?
14. Energy at n = 6, E = 
13.6
36
?
= –0.3777777
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
= 1.507777 = 
2
2 2
13.6 z 13.6
n n
?
?
or, n = 
13.6
1.507
= 3.03 = 3 = n.
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.
–0.85 eV
–1.5 eV
–3.4 eV
–13.6 eV
Bohr’s Theory and Physics of Atom
43.4
16. Energy in ground state is the energy acquired in the transition of 2
nd
excited state to ground state.
As 2
nd
excited state is taken as zero level.
E = 
15 8
9
1
hc 4.14 10 3 10 1242
46 46 10
?
?
? ? ?
? ?
? ?
= 27 ev.
Again energy in the first excited state
E = 
15 8
II
hc 4.14 10 3 10
103.5
?
? ? ?
?
?
= 12 ev.
17. a) The gas emits 6 wavelengths, let it be in nth excited state.
?
n(n 1)
2
?
= 6 ? n = 4 ? The gas is in 4
th
excited state.
b) Total no.of wavelengths in the transition is 6. We have 
n(n 1)
2
?
= 6 ? n = 4.
18. a) We know, m ? r = 
nh
2 ?
? mr
2
w = 
nh
2 ?
? w = 
2
hn
2 m r ? ? ?
= 
34
31 2 20
1 6.63 10
2 3.14 9.1 10 (0.53) 10
?
? ?
? ?
? ? ? ? ?
= 0.413 ? 10
17
rad/s = 4.13 ? 10
17
rad/s.
19. The range of Balmer series is 656.3 nm to 365 nm. It can resolve ? and ? + ?? if ?/ ?? = 8000.
? No.of wavelengths in the range = 
656.3 365
8000
?
= 36
Total no.of lines 36 + 2 = 38 [extra two is for first and last wavelength]
20. a) n
1
= 1, n
2
= 3, E = 13.6 (1/1 – 1/9) = 13.6 ? 8/9 = hc/ ?
or, 
15 8 7
13.6 8 4.14 10 3 10 4.14 3 10
9 13.6 8
? ?
? ? ? ? ? ?
? ? ? ?
? ?
= 1.027 ? 10
–7
= 103 nm.
b) As ‘n’ changes by 2, we may consider n = 2 to n = 4
then E = 13.6 ? (1/4 – 1/16) = 2.55 ev and 2.55 = 
1242
?
or ? = 487 nm. ?
21. Frequency of the revolution in the ground state is 
0
0
V
2 r ?
[r
0
= radius of ground state, V
0
= velocity in the ground state]
?Frequency of radiation emitted is 
0
0
V
2 r ?
= f
? C = f ? ? ? = C/f = 
0
0
C2 r
V
?
? ? = 
0
0
C2 r
V
?
= 45.686 nm = 45.7 nm. ?
22. KE = 3/2 KT = 1.5 KT, K = 8.62 ? 10
–5
eV/k, Binding Energy = –13.6 (1/ ? – 1/1) = 13.6 eV.
According to the question, 1.5 KT = 13.6
? 1.5 ? 8.62 ? 10
–5
? T = 13.6
? T = 
5
13.6
1.5 8.62 10
?
? ?
= 1.05 ? 10
5
K
No, because the molecule exists an H
2
+
which is impossible.
23. K = 8.62 ? 10
–5
eV/k
K.E. of H
2
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)
? 3/2 ? 8.62 ? 10
–5
T = 
13.6 8
9
?
? T = 0.9349 ? 10
5
= 9.349 ? 10
4
= 9.4 ? 10
4
K.
Bohr’s Theory and Physics of Atom
43.5
24. n = 2, T = 10
–8
s
Frequency = 
4
2 3 3
0
me
4 n h ?
So, time period = 1/f = 
2 3 3
4
4 o n h
me
?
?
2 3 3 24 102
4 76
4 (8.85) 2 (6.63) 10 10
9.1 (1.6) 10
? ?
?
? ? ? ?
?
?
= 12247.735 ? 10
–19
sec.
No.of revolutions = 
8
19
10
12247.735 10
?
?
?
= 8.16 ? 10
5
= 8.2 ? 10
6
revolution.
25. Dipole moment ( ?)
= n i A = 1 ? q/t A = qfA
=
5 2 2 4
2 2 0
0
2 3 3 2 3 3
0 0
me ( r n ) me
e ( r n )
4 h n 4 h n
? ?
? ? ? ?
? ?
=
31 19 5 2 20
12 2 34 3 3
(9.1 10 )(1.6 10 ) (0.53) 10 1
4 (8.85 10 ) (6.64 10 ) (1)
? ? ?
? ?
? ? ? ? ? ? ?
? ? ?
= 0.0009176 ? 10
–20
= 9.176 ? 10
–24
A - m
2
.
26. Magnetic Dipole moment = n i A = 
4 2 2
n
2 3 3
0
e me r n
4 h n
? ? ?
?
Angular momentum = mvr = 
nh
2 ?
Since the ratio of magnetic dipole moment and angular momentum is independent of Z.
Hence it is an universal constant.
Ratio = 
5 2 2
0
3 3
0
e m r n 2
nh 24 h n
? ? ? ?
?
?
?
19 5 31 2 10 2
12 2 34 4 2
(1.6 10 ) (9.1 10 ) (3.14) (0.53 10 )
2 (8.85 10 ) (6.63 10 ) 1
? ? ?
? ?
? ? ? ? ? ?
? ? ? ? ?
= 8.73 ? 10
10
C/kg.
27. The energies associated with 450 nm radiation = 
1242
450
= 2.76 ev
Energy associated with 550 nm radiation = 
1242
550
= 2.258 = 2.26 ev.
The light comes under visible range
Thus, n
1
= 2, n
2
= 3, 4, 5, ……
E
2
– E
3
= 13.6 (1/2
2
– 1/3
2
) = 1.9 ev
E
2
– E
4
= 13.6 (1/4 – 1/16) = 2.55 ev
E
2
– E
5
= 13.6 (1/4 – 1/25) = 2.856 ev
Only E
2
– E
4
comes in the range of energy provided. So the wavelength corresponding to that energy 
will be absorbed.
? = 
1242
2.55
= 487.05 nm = 487 nm
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
E
1
= 4 ? 13.6 (1 – 1/4) = 40.8 eV [E
1
> E hence it is not possible]
n = 1 to n = 3
E
2
= 4 ? 13.6 (1 – 1/9) = 48.3 eV [E
2
> E hence impossible]
Similarly n = 1 to n = 4 is also not possible.
n = 2 to n = 3
E
3
= 4 ? 13.6 (1/4 – 1/9) = 7.56 eV
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FAQs on HC Verma Solutions: Chapter 43 - Bohr's Model & Physics of Atom - Physics Class 11 - NEET

1. What is Bohr's model of the atom?
Ans. Bohr's model of the atom, proposed by Niels Bohr in 1913, describes the structure of an atom based on quantized energy levels. According to this model, electrons orbit the nucleus in specific circular paths or orbits. Each orbit corresponds to a specific energy level, and electrons can transition between these energy levels by absorbing or emitting energy in the form of photons.
2. How does Bohr's model explain the stability of atoms?
Ans. Bohr's model explains the stability of atoms by introducing the concept of stationary states. According to this model, electrons can only exist in specific energy levels or orbits around the nucleus. These energy levels are stable because electrons do not emit energy while in these stationary states. When an electron absorbs or emits energy, it transitions between energy levels, but it remains stable in each energy level.
3. What is the significance of the quantization of energy levels in Bohr's model?
Ans. The quantization of energy levels in Bohr's model is significant because it explains the discrete nature of atomic spectra. The energy levels in Bohr's model are quantized, meaning they can only take on specific values. This quantization leads to the observation of discrete lines in the atomic spectra, as each transition between energy levels corresponds to the emission or absorption of a specific wavelength of light.
4. How does Bohr's model explain the emission spectra of atoms?
Ans. Bohr's model explains the emission spectra of atoms by considering the transitions of electrons between energy levels. When an electron transitions from a higher energy level to a lower energy level, it emits energy in the form of a photon. The energy of the emitted photon corresponds to the energy difference between the initial and final energy levels. The emitted photons create a pattern of spectral lines, which can be used to identify the elements present in a sample.
5. What are the limitations of Bohr's model of the atom?
Ans. Although Bohr's model was a significant step in understanding the structure of atoms, it has some limitations. One limitation is that it fails to explain the behavior of multi-electron atoms, as it assumes a simplistic model with only one electron. Additionally, it cannot account for the concept of electron spin or the shapes of atomic orbitals. These limitations were later addressed by the development of more advanced quantum mechanical models of the atom.
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Document Description: HC Verma Solutions: Chapter 43 - Bohr's Model & Physics of Atom for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 43 - Bohr's Model & Physics of Atom have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 43 - Bohr's Model & Physics of Atom covers topics like and HC Verma Solutions: Chapter 43 - Bohr's Model & Physics of Atom Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 43 - Bohr's Model & Physics of Atom.
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