Solved Example: Dual Nature of Matter and Radiation | Physics for JEE Main & Advanced PDF Download

 Page 1


JEE Solved Example on Dual Nature of Matter and 
Radiation 
JEE Mains 
Q1: What is the de-Broglie wavelength of the ?? -particle accelerated through a potential difference 
?? 
(A) 
?? .??????
v ?? Å 
(B) 
???? .????
v ?? Å 
(C) 
?? .??????
v ?? ?? 
(D) 
?? .??????
v ?? ?? 
Ans: (C) 1 =
h
v 2????
=
h
v2?? 2
?? 2
?? 
On putting ?? ?? = 2
'
1.6
'
10
-19
C 
?? a
= 4 m
p
= 4
'
1.67 · 10
-27
 kg p 1 =
0.101
v ?? A 
Q2: The de-Broglie wavelength of a particle moving with a velocity ?? . ???? · ????
?? ?? /?? is equal to the 
wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is 
(velocity of light is ?? . ????
?? ?? /?? ) 
(A) ?? /?? 
(B) ?? /?? 
(C) ?? /?? 
(D) ?? /?? 
Ans: (B) ?? partice 
=
1
2
?? ?? 2
 also 1 =
h
????
 
b ?? partice 
=
1
2
Rh
?? ?
¨
F
?
× ?? 2
=
?? h
21
 
b K
pholon 
=
hc
1
 
K
partide 
K
protoon 
=
V
2C
=
2.25 · 10
8
2
'
3
'
10
8
=
3
8
 
Q3: The kinetic energy of electron and proton is ????
-????
 ?? . Then the relation between their de -
Broglie wavelengths is 
(A) ?? ?? < ?? ? 
(B) ?? ?? > ?? ? 
(C) ?? ?? = ?? ? 
(D) ?? ?? = ???? 
Ans: (A) By using 1 =
h
v 2mE
 E = 10
-32
 J 3/4 Constant for both particles. Hence 1?? 1
v ?? Since ?? ?? >
?? ?? so 1
?? < 1
?? . 
 
Page 2


JEE Solved Example on Dual Nature of Matter and 
Radiation 
JEE Mains 
Q1: What is the de-Broglie wavelength of the ?? -particle accelerated through a potential difference 
?? 
(A) 
?? .??????
v ?? Å 
(B) 
???? .????
v ?? Å 
(C) 
?? .??????
v ?? ?? 
(D) 
?? .??????
v ?? ?? 
Ans: (C) 1 =
h
v 2????
=
h
v2?? 2
?? 2
?? 
On putting ?? ?? = 2
'
1.6
'
10
-19
C 
?? a
= 4 m
p
= 4
'
1.67 · 10
-27
 kg p 1 =
0.101
v ?? A 
Q2: The de-Broglie wavelength of a particle moving with a velocity ?? . ???? · ????
?? ?? /?? is equal to the 
wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is 
(velocity of light is ?? . ????
?? ?? /?? ) 
(A) ?? /?? 
(B) ?? /?? 
(C) ?? /?? 
(D) ?? /?? 
Ans: (B) ?? partice 
=
1
2
?? ?? 2
 also 1 =
h
????
 
b ?? partice 
=
1
2
Rh
?? ?
¨
F
?
× ?? 2
=
?? h
21
 
b K
pholon 
=
hc
1
 
K
partide 
K
protoon 
=
V
2C
=
2.25 · 10
8
2
'
3
'
10
8
=
3
8
 
Q3: The kinetic energy of electron and proton is ????
-????
 ?? . Then the relation between their de -
Broglie wavelengths is 
(A) ?? ?? < ?? ? 
(B) ?? ?? > ?? ? 
(C) ?? ?? = ?? ? 
(D) ?? ?? = ???? 
Ans: (A) By using 1 =
h
v 2mE
 E = 10
-32
 J 3/4 Constant for both particles. Hence 1?? 1
v ?? Since ?? ?? >
?? ?? so 1
?? < 1
?? . 
 
Q4: The de-Broglie wavelength of a particle accelerated with 150 volt potential is ????
-????
 ?? . If it is 
accelerated by 600 volts p.d., its wavelength will be 
(A) ?? . ???? Å 
(B) ?? . ?? Å 
(C) ?? . ?? Å 
(D) ?? Å 
Ans: (B) By using 1?? 1
v ?? P 
1
1
1
2
= v
?? 2
?? 1
 
b 
10
-10
1
2
= v
600
150
= 2 b  1
2
= 0.5Å 
Q5: When the momentum of a proton is changed by an amount ?? ?? , the corresponding change in 
the de-Broglie wavelength is found to be ?? . ???? %. Then, the original momentum of the proton was 
(A) ?? ?? 
(B) ?????? ?? ?? 
(C) ?????? ?? ?? 
(D) ?? ?? ?? 
Ans: (C) 1?? 1
p
  p  
D?? p
= -
D
1
  p  |
Dp
p
| = |
D
1
| 
b 
?? 0
p
=
0.25
100
=
1
400
  p  p = 400 p
0
. 
Q6: The work function of metal is ?????? . Light of wavelength ???????? Å is incident on this metal 
surface. The velocity of emitted photo -electrons will be 
(A) ???? ?? /?????? 
(B) ?? · ????
?? ?? /?????? 
(C) ?? · ????
?? ?? /?????? 
(D) ?? · ????
?? ?? /?????? 
Ans: (D) ?? = ?? 0
+ ?? max
; ?? =
12375
3000
= 4.125eV 
P K
max 
= E - W
0
= 4.125eV- 1eV= 3.125eV 
P  
1
2
?? ?? max
2
= 3.125· 1.6 · 10
-19
 J 
p  V
max
= v
2·3.125·1.6·10
-19
9.1·10
-31
= 1 · 10
6
 m /s 
 
Q7: The work function of a metallic surface is ?? . ???????? . The photo-electrons are emitted when light 
of wavelength 2000 Å falls on it. The potential difference applied to stop the fastest photo -
electrons is [?? = ?? . ????
'
????
-????
?????????? ] 
(A) 1.2 volts 
(B) 2.24 volts 
(C) 3.6 volts 
(D) 4.8 volts 
Ans: (A) Energy of incident light ?? =
12375
2000
= 6.18eV 
According to relation E = W
0
+ eV
0
 
 p  ?? 0
=
(?? - ?? 0
)
?? =
(6.18eV- 5.01eV )
e
= 1.17 V » 1.2 V 
Page 3


JEE Solved Example on Dual Nature of Matter and 
Radiation 
JEE Mains 
Q1: What is the de-Broglie wavelength of the ?? -particle accelerated through a potential difference 
?? 
(A) 
?? .??????
v ?? Å 
(B) 
???? .????
v ?? Å 
(C) 
?? .??????
v ?? ?? 
(D) 
?? .??????
v ?? ?? 
Ans: (C) 1 =
h
v 2????
=
h
v2?? 2
?? 2
?? 
On putting ?? ?? = 2
'
1.6
'
10
-19
C 
?? a
= 4 m
p
= 4
'
1.67 · 10
-27
 kg p 1 =
0.101
v ?? A 
Q2: The de-Broglie wavelength of a particle moving with a velocity ?? . ???? · ????
?? ?? /?? is equal to the 
wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is 
(velocity of light is ?? . ????
?? ?? /?? ) 
(A) ?? /?? 
(B) ?? /?? 
(C) ?? /?? 
(D) ?? /?? 
Ans: (B) ?? partice 
=
1
2
?? ?? 2
 also 1 =
h
????
 
b ?? partice 
=
1
2
Rh
?? ?
¨
F
?
× ?? 2
=
?? h
21
 
b K
pholon 
=
hc
1
 
K
partide 
K
protoon 
=
V
2C
=
2.25 · 10
8
2
'
3
'
10
8
=
3
8
 
Q3: The kinetic energy of electron and proton is ????
-????
 ?? . Then the relation between their de -
Broglie wavelengths is 
(A) ?? ?? < ?? ? 
(B) ?? ?? > ?? ? 
(C) ?? ?? = ?? ? 
(D) ?? ?? = ???? 
Ans: (A) By using 1 =
h
v 2mE
 E = 10
-32
 J 3/4 Constant for both particles. Hence 1?? 1
v ?? Since ?? ?? >
?? ?? so 1
?? < 1
?? . 
 
Q4: The de-Broglie wavelength of a particle accelerated with 150 volt potential is ????
-????
 ?? . If it is 
accelerated by 600 volts p.d., its wavelength will be 
(A) ?? . ???? Å 
(B) ?? . ?? Å 
(C) ?? . ?? Å 
(D) ?? Å 
Ans: (B) By using 1?? 1
v ?? P 
1
1
1
2
= v
?? 2
?? 1
 
b 
10
-10
1
2
= v
600
150
= 2 b  1
2
= 0.5Å 
Q5: When the momentum of a proton is changed by an amount ?? ?? , the corresponding change in 
the de-Broglie wavelength is found to be ?? . ???? %. Then, the original momentum of the proton was 
(A) ?? ?? 
(B) ?????? ?? ?? 
(C) ?????? ?? ?? 
(D) ?? ?? ?? 
Ans: (C) 1?? 1
p
  p  
D?? p
= -
D
1
  p  |
Dp
p
| = |
D
1
| 
b 
?? 0
p
=
0.25
100
=
1
400
  p  p = 400 p
0
. 
Q6: The work function of metal is ?????? . Light of wavelength ???????? Å is incident on this metal 
surface. The velocity of emitted photo -electrons will be 
(A) ???? ?? /?????? 
(B) ?? · ????
?? ?? /?????? 
(C) ?? · ????
?? ?? /?????? 
(D) ?? · ????
?? ?? /?????? 
Ans: (D) ?? = ?? 0
+ ?? max
; ?? =
12375
3000
= 4.125eV 
P K
max 
= E - W
0
= 4.125eV- 1eV= 3.125eV 
P  
1
2
?? ?? max
2
= 3.125· 1.6 · 10
-19
 J 
p  V
max
= v
2·3.125·1.6·10
-19
9.1·10
-31
= 1 · 10
6
 m /s 
 
Q7: The work function of a metallic surface is ?? . ???????? . The photo-electrons are emitted when light 
of wavelength 2000 Å falls on it. The potential difference applied to stop the fastest photo -
electrons is [?? = ?? . ????
'
????
-????
?????????? ] 
(A) 1.2 volts 
(B) 2.24 volts 
(C) 3.6 volts 
(D) 4.8 volts 
Ans: (A) Energy of incident light ?? =
12375
2000
= 6.18eV 
According to relation E = W
0
+ eV
0
 
 p  ?? 0
=
(?? - ?? 0
)
?? =
(6.18eV- 5.01eV )
e
= 1.17 V » 1.2 V 
Q8: A particle of mass ?? at rest decays into two particles of masses ?? ?? and ?? ?? , having non-zero 
velocities. The ratio of the de -Broglie wavelengths of the particles, ?? ?? /?? ?? is 
(A) ?? ?? /?? ?? 
(B) ?? ?? /?? ?? 
(C) 1.0 
(D) 
v
?? ?? /
v
?? ?? 
Ans: (C) By law of conservation of momentum 
0 = ?? 1
?? ?
1
+ ?? 2
?? ?
2
  b  ?? 1
?? ?
1
= -?? 2
?? ?
2
 
-ve sign indicates that both the particles are moving in opposite direction. Now de -Broglie 
wavelengths 
1
1
=
h
?? 1
?? 1
  and  1
2
=
h
?? 2
?? 2
; 
1
1
1
2
=
?? 2
?? 2
?? 1
?? 1
= 1 
Q9: When photon of energy ?? . ???????? strike the surface of a metal ?? , the ejected photoelectrons 
have maximum kinetic energy ?? ?? ???? and de-Broglie wavelength ?? ?? . The maximum kinetic energy 
of photoelectrons liberated from another metal B by photon of energy ?? . ???????? is ?? ?? =
(?? ?? - ?? . ???? )???? . If the de-Broglie wavelength of these photoelectrons is ?? ?? = ????
?? , then 
(A) The work function of ?? is ?? . ???????? 
(B) The work function of ?? is ?? . ???????? 
(C) ?? ?? = ?? . ???????? 
(D) ?? ?? = ?? . ???????? 
Ans: (A, B, C) K
max
= E - W
0
 
?? ?? = 4.25 - (?? 0
)
?? ?? ?? = (?? ?? - 1.5) = 4.70 - (?? 0
)
?? 
Equation (i) and (ii) gives (W
0
)
B
= (W
0
)
A
= 1.95eV 
De Broglie wavelength 1 =
h
v 2mK
, 1?? 1
v ?? 
p 
1
?? 1
?? = v
?? ?? ?? ?? p 2 = v
?? ?? ?? ?? -1.5
 p  ?? ?? = 2eV 
From equation (i) and (ii) 
W
A
= 2.25eV and W
B
= 4.20eV 
Q10: In a photoemissive cell with executing wavelength ?? , the fastest electron has speed ?? . If the 
exciting wavelength is changed to ?? ?? /?? , the speed of the fastest emitted electron will be 
(A) ?? (?? /?? )
?? /?? 
(B) ?? (?? /?? )
?? /?? 
(C) Less than ?? (?? /?? )
?? /?? 
(D) Greater than ?? (?? /?? )
?? /?? 
Ans:  
Page 4


JEE Solved Example on Dual Nature of Matter and 
Radiation 
JEE Mains 
Q1: What is the de-Broglie wavelength of the ?? -particle accelerated through a potential difference 
?? 
(A) 
?? .??????
v ?? Å 
(B) 
???? .????
v ?? Å 
(C) 
?? .??????
v ?? ?? 
(D) 
?? .??????
v ?? ?? 
Ans: (C) 1 =
h
v 2????
=
h
v2?? 2
?? 2
?? 
On putting ?? ?? = 2
'
1.6
'
10
-19
C 
?? a
= 4 m
p
= 4
'
1.67 · 10
-27
 kg p 1 =
0.101
v ?? A 
Q2: The de-Broglie wavelength of a particle moving with a velocity ?? . ???? · ????
?? ?? /?? is equal to the 
wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is 
(velocity of light is ?? . ????
?? ?? /?? ) 
(A) ?? /?? 
(B) ?? /?? 
(C) ?? /?? 
(D) ?? /?? 
Ans: (B) ?? partice 
=
1
2
?? ?? 2
 also 1 =
h
????
 
b ?? partice 
=
1
2
Rh
?? ?
¨
F
?
× ?? 2
=
?? h
21
 
b K
pholon 
=
hc
1
 
K
partide 
K
protoon 
=
V
2C
=
2.25 · 10
8
2
'
3
'
10
8
=
3
8
 
Q3: The kinetic energy of electron and proton is ????
-????
 ?? . Then the relation between their de -
Broglie wavelengths is 
(A) ?? ?? < ?? ? 
(B) ?? ?? > ?? ? 
(C) ?? ?? = ?? ? 
(D) ?? ?? = ???? 
Ans: (A) By using 1 =
h
v 2mE
 E = 10
-32
 J 3/4 Constant for both particles. Hence 1?? 1
v ?? Since ?? ?? >
?? ?? so 1
?? < 1
?? . 
 
Q4: The de-Broglie wavelength of a particle accelerated with 150 volt potential is ????
-????
 ?? . If it is 
accelerated by 600 volts p.d., its wavelength will be 
(A) ?? . ???? Å 
(B) ?? . ?? Å 
(C) ?? . ?? Å 
(D) ?? Å 
Ans: (B) By using 1?? 1
v ?? P 
1
1
1
2
= v
?? 2
?? 1
 
b 
10
-10
1
2
= v
600
150
= 2 b  1
2
= 0.5Å 
Q5: When the momentum of a proton is changed by an amount ?? ?? , the corresponding change in 
the de-Broglie wavelength is found to be ?? . ???? %. Then, the original momentum of the proton was 
(A) ?? ?? 
(B) ?????? ?? ?? 
(C) ?????? ?? ?? 
(D) ?? ?? ?? 
Ans: (C) 1?? 1
p
  p  
D?? p
= -
D
1
  p  |
Dp
p
| = |
D
1
| 
b 
?? 0
p
=
0.25
100
=
1
400
  p  p = 400 p
0
. 
Q6: The work function of metal is ?????? . Light of wavelength ???????? Å is incident on this metal 
surface. The velocity of emitted photo -electrons will be 
(A) ???? ?? /?????? 
(B) ?? · ????
?? ?? /?????? 
(C) ?? · ????
?? ?? /?????? 
(D) ?? · ????
?? ?? /?????? 
Ans: (D) ?? = ?? 0
+ ?? max
; ?? =
12375
3000
= 4.125eV 
P K
max 
= E - W
0
= 4.125eV- 1eV= 3.125eV 
P  
1
2
?? ?? max
2
= 3.125· 1.6 · 10
-19
 J 
p  V
max
= v
2·3.125·1.6·10
-19
9.1·10
-31
= 1 · 10
6
 m /s 
 
Q7: The work function of a metallic surface is ?? . ???????? . The photo-electrons are emitted when light 
of wavelength 2000 Å falls on it. The potential difference applied to stop the fastest photo -
electrons is [?? = ?? . ????
'
????
-????
?????????? ] 
(A) 1.2 volts 
(B) 2.24 volts 
(C) 3.6 volts 
(D) 4.8 volts 
Ans: (A) Energy of incident light ?? =
12375
2000
= 6.18eV 
According to relation E = W
0
+ eV
0
 
 p  ?? 0
=
(?? - ?? 0
)
?? =
(6.18eV- 5.01eV )
e
= 1.17 V » 1.2 V 
Q8: A particle of mass ?? at rest decays into two particles of masses ?? ?? and ?? ?? , having non-zero 
velocities. The ratio of the de -Broglie wavelengths of the particles, ?? ?? /?? ?? is 
(A) ?? ?? /?? ?? 
(B) ?? ?? /?? ?? 
(C) 1.0 
(D) 
v
?? ?? /
v
?? ?? 
Ans: (C) By law of conservation of momentum 
0 = ?? 1
?? ?
1
+ ?? 2
?? ?
2
  b  ?? 1
?? ?
1
= -?? 2
?? ?
2
 
-ve sign indicates that both the particles are moving in opposite direction. Now de -Broglie 
wavelengths 
1
1
=
h
?? 1
?? 1
  and  1
2
=
h
?? 2
?? 2
; 
1
1
1
2
=
?? 2
?? 2
?? 1
?? 1
= 1 
Q9: When photon of energy ?? . ???????? strike the surface of a metal ?? , the ejected photoelectrons 
have maximum kinetic energy ?? ?? ???? and de-Broglie wavelength ?? ?? . The maximum kinetic energy 
of photoelectrons liberated from another metal B by photon of energy ?? . ???????? is ?? ?? =
(?? ?? - ?? . ???? )???? . If the de-Broglie wavelength of these photoelectrons is ?? ?? = ????
?? , then 
(A) The work function of ?? is ?? . ???????? 
(B) The work function of ?? is ?? . ???????? 
(C) ?? ?? = ?? . ???????? 
(D) ?? ?? = ?? . ???????? 
Ans: (A, B, C) K
max
= E - W
0
 
?? ?? = 4.25 - (?? 0
)
?? ?? ?? = (?? ?? - 1.5) = 4.70 - (?? 0
)
?? 
Equation (i) and (ii) gives (W
0
)
B
= (W
0
)
A
= 1.95eV 
De Broglie wavelength 1 =
h
v 2mK
, 1?? 1
v ?? 
p 
1
?? 1
?? = v
?? ?? ?? ?? p 2 = v
?? ?? ?? ?? -1.5
 p  ?? ?? = 2eV 
From equation (i) and (ii) 
W
A
= 2.25eV and W
B
= 4.20eV 
Q10: In a photoemissive cell with executing wavelength ?? , the fastest electron has speed ?? . If the 
exciting wavelength is changed to ?? ?? /?? , the speed of the fastest emitted electron will be 
(A) ?? (?? /?? )
?? /?? 
(B) ?? (?? /?? )
?? /?? 
(C) Less than ?? (?? /?? )
?? /?? 
(D) Greater than ?? (?? /?? )
?? /?? 
Ans:  
 (D) hv - W
0
=
1
2
?? ?? max
2
?
h?? ?? -
h?? ?? 0
=
1
2
?? ?? max
2
 ? h?? (
?? 0
- ?? ?? ?? 0
) =
1
2
?? ?? max
2
? ?? max
= v
2h?? ?? (
?? 0
- ?? ?? ?? 0
)
 
(D) h?? - ?? 0
=
1
2
?? ?? max
2
?
h?? ?? -
h?? ?? 0
=
1
2
?? ?? max
2
 
When wavelength is ?? and velocity is ?? , then 
?? = v
2hc
m
(
?? 0
- ?? ?? ?? 0
) 
When wavelength is 
3?? 4
 and velocity is ?? '
 then 
?? '
= v
2?? h
?? [
?? 0
- 3?? /4
(3?? /4) × ?? 0
] 
Divide equation (ii) by (i), we get 
?? '
?? =
v
?? 0
- (3?? /4)]
3
4
?? ?? 0
×
?? ?? 0
?? 0
- ?? ?? '
= ?? (
4
3
)
4/2
v
[?? 0
- (3?? /4)]
?? 0
- ?? i.e. ?? '
> ?? (
4
3
)
1/2
 
Q11: Photoelectric emission is observed from a metallic surface for frequencies ?? ?? and ?? ?? of the 
incident light rays (?? ?? > ?? ?? ). If the maximum values of kinetic energy of the photoelectrons 
emitted in the two cases are in the ratio of 1 : ?? , then the threshold frequency of the metallic 
surface is 
(A) 
?? ?? -?? ?? ?? -?? 
(B) 
?? ?? ?? -?? ?? ?? -?? 
(C) 
?? ?? ?? -?? ?? ?? -?? 
(D) 
?? ?? -?? ?? ?? 
Ans: (B) By using h?? - h?? 0
= ?? max
 
? h(v
1
- v
0
) = k
1
 
And h(?? 2
- ?? 0
) = ?? 2
 
?
?? 1
-?? 0
?? 2
-?? 0
=
?? 1
?? 2
=
1
?? , 
Hence ?? 0
=
?? ?? 1
-?? 2
?? -1
. 
An X-ray tube is operating at 50kV and 20 mA . The target material of the tube has a mass of 1.0 kg 
and specific heat 495 J kg
-10
C
-1
. One percent of the supplied electric power is converted into X-
Page 5


JEE Solved Example on Dual Nature of Matter and 
Radiation 
JEE Mains 
Q1: What is the de-Broglie wavelength of the ?? -particle accelerated through a potential difference 
?? 
(A) 
?? .??????
v ?? Å 
(B) 
???? .????
v ?? Å 
(C) 
?? .??????
v ?? ?? 
(D) 
?? .??????
v ?? ?? 
Ans: (C) 1 =
h
v 2????
=
h
v2?? 2
?? 2
?? 
On putting ?? ?? = 2
'
1.6
'
10
-19
C 
?? a
= 4 m
p
= 4
'
1.67 · 10
-27
 kg p 1 =
0.101
v ?? A 
Q2: The de-Broglie wavelength of a particle moving with a velocity ?? . ???? · ????
?? ?? /?? is equal to the 
wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is 
(velocity of light is ?? . ????
?? ?? /?? ) 
(A) ?? /?? 
(B) ?? /?? 
(C) ?? /?? 
(D) ?? /?? 
Ans: (B) ?? partice 
=
1
2
?? ?? 2
 also 1 =
h
????
 
b ?? partice 
=
1
2
Rh
?? ?
¨
F
?
× ?? 2
=
?? h
21
 
b K
pholon 
=
hc
1
 
K
partide 
K
protoon 
=
V
2C
=
2.25 · 10
8
2
'
3
'
10
8
=
3
8
 
Q3: The kinetic energy of electron and proton is ????
-????
 ?? . Then the relation between their de -
Broglie wavelengths is 
(A) ?? ?? < ?? ? 
(B) ?? ?? > ?? ? 
(C) ?? ?? = ?? ? 
(D) ?? ?? = ???? 
Ans: (A) By using 1 =
h
v 2mE
 E = 10
-32
 J 3/4 Constant for both particles. Hence 1?? 1
v ?? Since ?? ?? >
?? ?? so 1
?? < 1
?? . 
 
Q4: The de-Broglie wavelength of a particle accelerated with 150 volt potential is ????
-????
 ?? . If it is 
accelerated by 600 volts p.d., its wavelength will be 
(A) ?? . ???? Å 
(B) ?? . ?? Å 
(C) ?? . ?? Å 
(D) ?? Å 
Ans: (B) By using 1?? 1
v ?? P 
1
1
1
2
= v
?? 2
?? 1
 
b 
10
-10
1
2
= v
600
150
= 2 b  1
2
= 0.5Å 
Q5: When the momentum of a proton is changed by an amount ?? ?? , the corresponding change in 
the de-Broglie wavelength is found to be ?? . ???? %. Then, the original momentum of the proton was 
(A) ?? ?? 
(B) ?????? ?? ?? 
(C) ?????? ?? ?? 
(D) ?? ?? ?? 
Ans: (C) 1?? 1
p
  p  
D?? p
= -
D
1
  p  |
Dp
p
| = |
D
1
| 
b 
?? 0
p
=
0.25
100
=
1
400
  p  p = 400 p
0
. 
Q6: The work function of metal is ?????? . Light of wavelength ???????? Å is incident on this metal 
surface. The velocity of emitted photo -electrons will be 
(A) ???? ?? /?????? 
(B) ?? · ????
?? ?? /?????? 
(C) ?? · ????
?? ?? /?????? 
(D) ?? · ????
?? ?? /?????? 
Ans: (D) ?? = ?? 0
+ ?? max
; ?? =
12375
3000
= 4.125eV 
P K
max 
= E - W
0
= 4.125eV- 1eV= 3.125eV 
P  
1
2
?? ?? max
2
= 3.125· 1.6 · 10
-19
 J 
p  V
max
= v
2·3.125·1.6·10
-19
9.1·10
-31
= 1 · 10
6
 m /s 
 
Q7: The work function of a metallic surface is ?? . ???????? . The photo-electrons are emitted when light 
of wavelength 2000 Å falls on it. The potential difference applied to stop the fastest photo -
electrons is [?? = ?? . ????
'
????
-????
?????????? ] 
(A) 1.2 volts 
(B) 2.24 volts 
(C) 3.6 volts 
(D) 4.8 volts 
Ans: (A) Energy of incident light ?? =
12375
2000
= 6.18eV 
According to relation E = W
0
+ eV
0
 
 p  ?? 0
=
(?? - ?? 0
)
?? =
(6.18eV- 5.01eV )
e
= 1.17 V » 1.2 V 
Q8: A particle of mass ?? at rest decays into two particles of masses ?? ?? and ?? ?? , having non-zero 
velocities. The ratio of the de -Broglie wavelengths of the particles, ?? ?? /?? ?? is 
(A) ?? ?? /?? ?? 
(B) ?? ?? /?? ?? 
(C) 1.0 
(D) 
v
?? ?? /
v
?? ?? 
Ans: (C) By law of conservation of momentum 
0 = ?? 1
?? ?
1
+ ?? 2
?? ?
2
  b  ?? 1
?? ?
1
= -?? 2
?? ?
2
 
-ve sign indicates that both the particles are moving in opposite direction. Now de -Broglie 
wavelengths 
1
1
=
h
?? 1
?? 1
  and  1
2
=
h
?? 2
?? 2
; 
1
1
1
2
=
?? 2
?? 2
?? 1
?? 1
= 1 
Q9: When photon of energy ?? . ???????? strike the surface of a metal ?? , the ejected photoelectrons 
have maximum kinetic energy ?? ?? ???? and de-Broglie wavelength ?? ?? . The maximum kinetic energy 
of photoelectrons liberated from another metal B by photon of energy ?? . ???????? is ?? ?? =
(?? ?? - ?? . ???? )???? . If the de-Broglie wavelength of these photoelectrons is ?? ?? = ????
?? , then 
(A) The work function of ?? is ?? . ???????? 
(B) The work function of ?? is ?? . ???????? 
(C) ?? ?? = ?? . ???????? 
(D) ?? ?? = ?? . ???????? 
Ans: (A, B, C) K
max
= E - W
0
 
?? ?? = 4.25 - (?? 0
)
?? ?? ?? = (?? ?? - 1.5) = 4.70 - (?? 0
)
?? 
Equation (i) and (ii) gives (W
0
)
B
= (W
0
)
A
= 1.95eV 
De Broglie wavelength 1 =
h
v 2mK
, 1?? 1
v ?? 
p 
1
?? 1
?? = v
?? ?? ?? ?? p 2 = v
?? ?? ?? ?? -1.5
 p  ?? ?? = 2eV 
From equation (i) and (ii) 
W
A
= 2.25eV and W
B
= 4.20eV 
Q10: In a photoemissive cell with executing wavelength ?? , the fastest electron has speed ?? . If the 
exciting wavelength is changed to ?? ?? /?? , the speed of the fastest emitted electron will be 
(A) ?? (?? /?? )
?? /?? 
(B) ?? (?? /?? )
?? /?? 
(C) Less than ?? (?? /?? )
?? /?? 
(D) Greater than ?? (?? /?? )
?? /?? 
Ans:  
 (D) hv - W
0
=
1
2
?? ?? max
2
?
h?? ?? -
h?? ?? 0
=
1
2
?? ?? max
2
 ? h?? (
?? 0
- ?? ?? ?? 0
) =
1
2
?? ?? max
2
? ?? max
= v
2h?? ?? (
?? 0
- ?? ?? ?? 0
)
 
(D) h?? - ?? 0
=
1
2
?? ?? max
2
?
h?? ?? -
h?? ?? 0
=
1
2
?? ?? max
2
 
When wavelength is ?? and velocity is ?? , then 
?? = v
2hc
m
(
?? 0
- ?? ?? ?? 0
) 
When wavelength is 
3?? 4
 and velocity is ?? '
 then 
?? '
= v
2?? h
?? [
?? 0
- 3?? /4
(3?? /4) × ?? 0
] 
Divide equation (ii) by (i), we get 
?? '
?? =
v
?? 0
- (3?? /4)]
3
4
?? ?? 0
×
?? ?? 0
?? 0
- ?? ?? '
= ?? (
4
3
)
4/2
v
[?? 0
- (3?? /4)]
?? 0
- ?? i.e. ?? '
> ?? (
4
3
)
1/2
 
Q11: Photoelectric emission is observed from a metallic surface for frequencies ?? ?? and ?? ?? of the 
incident light rays (?? ?? > ?? ?? ). If the maximum values of kinetic energy of the photoelectrons 
emitted in the two cases are in the ratio of 1 : ?? , then the threshold frequency of the metallic 
surface is 
(A) 
?? ?? -?? ?? ?? -?? 
(B) 
?? ?? ?? -?? ?? ?? -?? 
(C) 
?? ?? ?? -?? ?? ?? -?? 
(D) 
?? ?? -?? ?? ?? 
Ans: (B) By using h?? - h?? 0
= ?? max
 
? h(v
1
- v
0
) = k
1
 
And h(?? 2
- ?? 0
) = ?? 2
 
?
?? 1
-?? 0
?? 2
-?? 0
=
?? 1
?? 2
=
1
?? , 
Hence ?? 0
=
?? ?? 1
-?? 2
?? -1
. 
An X-ray tube is operating at 50kV and 20 mA . The target material of the tube has a mass of 1.0 kg 
and specific heat 495 J kg
-10
C
-1
. One percent of the supplied electric power is converted into X-
Q12: rays and the entire remaining energy goes into heating the target. Then 
(A) A suitable target material must have a high melting temperature 
(B) A suitable target material must have low thermal conductivity 
(C) The average rate of rise of temperature of target would be ?? 
°
?? /?? 
(D) The minimum wavelength of the X-rays emitted is about ?? . ???? · ????
-????
 ?? 
Ans:  (A, C, D)P = VI = 50 · 10
3
· 20 · 10
-3
= 1000 W 
Power converted into heat 3/4990 W 
msDT= 990 b  DT = 2
°
C/sec 
Now 
h?? 1
min 
= ????  p  1
min 
=
h?? eV
= 0.248· 10
-10
 m 
Q13: Electrons with energy ?????????? are incident on the tungsten target of an X-ray tube. K shell 
electrons of tungsten have ionization energy ???? . ???????? . X-rays emitted by the tube contain only 
(A) A continuous X -ray spectrum (Bremsstrahlung) with a minimum wavelength of ~ ?? . ?????? Å. 
(B) A continuous X -ray spectrum (Bremsstrahlung) with all wavelengths. 
(C) The characteristic X -rays spectrum of tungsten. 
(D) A continuous X -ray spectrum (Bremsstrahlung) with a minimum wavelength of ~ ?? . ?????? Å and 
the characteristic X -ray spectrum of tungsten. 
Ans: (D) Minimum wavelength of continuous X -ray spectrum is given by 1
min
( in Å) =
12375
?? (???? )
=
12375
80
'
10
3
»0.155 
Q14: The ?? -ray wavelength of ?? ?? line of platinum (?? = ???? ) is 1.30 Å. The X-ray wavelength of ?? ?? 
line of Molybdenum (?? = ???? ) is 
(A) ?? . ???? Å 
(B) ?? . ???? Å 
(C) ?? . ???? Å 
(D) ?? . ???? Å 
Ans:  The wavelength of ?? ?? line is given by 
1
?? = ?? (?? - 7.4)
2
(
1
2
2
-
1
3
2
) ? ?? ?
1
(?? - 7.4)
2
 ?
?? 1
?? 2
=
(?? 2
- 7.4)
2
(?? 1
- 7.4)
2
?
1.30
?? 2
=
(4.2 - 7.4)
2
(78 - 7.4)
2
? ?? 2
= 5.41Å
 
Q15: The ratio of de -Broglie wavelengths of molecules of hydrogen and helium which are at 
temperature ????
°
?? and ??????
°
?? respectively is 
(A) 
?? ?? 
(B) v
?? ?? 
 (C) v
?? ?? 
(D) 1 
Ans: (C) de-Broglie wavelength ?? =
h
mv?? ms
, rms velocity of a gas particle at the given temperature (T) 
is given as