61. A 5 watt source emits monochromatic light of wavelength 5000 Å. When placed 0.5 m away, it liberates photo electrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, the number of photo electrons liberated will be reduced by a factor of

1. 1. 8

   
   
   

   2. 16

   
   
   

   3. 2

   
   
   

   4. 4

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Number of emitted electrons N_E ∝ Intensity
   

   ∝	1
   	(Distance)2

   
   Therefore, as distance is doubled, N_E decreases by (1/4) times.

   Correct Option: D

   Number of emitted electrons N_E ∝ Intensity
   

   ∝	1
   	(Distance)2

   
   Therefore, as distance is doubled, N_E decreases by (1/4) times.

---

62. When photons of energy hν fall on an aluminium plate (of work function E₀), photo electrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photo electrons will be

1. 1. 2K

   
   
   

   2. K

   
   
   

   3. K + hν

   
   
   

   4. K + E₀

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Applying Einstein's formula for photo-electricity
   

   hv = φ +	1	mv2 ;
   	2

   
   hv = φ + K
   φ = hv - K
   If we use 2ν frequency then let the kinetic energy becomes K' ​
   So,​h . 2ν = φ + K' ​
   2hν = hν – K + K' ​
   K' = hν + K

   Correct Option: C

   Applying Einstein's formula for photo-electricity
   

   hv = φ +	1	mv2 ;
   	2

   
   hv = φ + K
   φ = hv - K
   If we use 2ν frequency then let the kinetic energy becomes K' ​
   So,​h . 2ν = φ + K' ​
   2hν = hν – K + K' ​
   K' = hν + K

---

---

63. A photo-cell employs photoelectric effect to convert

1. 1. change in the intensity of illumination into a change in photoelectric current

   
   
   

   2. change in the intensity of illumination into a change in the work function of the photo cathode

   
   
   

   3. change in the frequency of light into a change in the electric current

   
   
   

   4. change in the frequency of light into a change in electric voltage

   
   
   

---

1. View Hint View Answer Discuss in Forum

   A photo-cell employs photoelectric effect to convert light energy into photoelectric current.

   Correct Option: A

   A photo-cell employs photoelectric effect to convert light energy into photoelectric current.

---

64. The momentum of a photon of energy 1 MeV in kg m/s, will be

1. 1. 7 × 10^–24

   
   
   

   2. 10^–22

   
   
   

   3. 5 × 10^–22

   
   
   

   4. 0.33 × 10⁶

   
   
   

---

1. View Hint View Answer Discuss in Forum

   1 MeV = 10⁶ × 1.6 × 10^–19 joule ​Momentum of photon
   

   =	E	=	1.6 × 10–13
   	c		3 × 108

   
   

   =	1.6	× 10–21 =	16	× 10–22
   	3		3

   
   = 5 × 10^–22 kg m/sec

   Correct Option: C

   1 MeV = 10⁶ × 1.6 × 10^–19 joule ​Momentum of photon
   

   =	E	=	1.6 × 10–13
   	c		3 × 108

   
   

   =	1.6	× 10–21 =	16	× 10–22
   	3		3

   
   = 5 × 10^–22 kg m/sec

---

---

65. A photosensitive metallic surface has work function, hν₀. If photons of energy 2 hν₀ fall on this surface, the electrons come out with a maximum velocity of 4 × 10⁶ m/s. When the photon energy is increased to 5 hν₀, then maximum velocity of photo electrons will be

1. 1. 2 × 10⁷ m/s

   
   
   

   2. 2 × 10⁶ m/s

   
   
   

   3. 8 × 10⁶ m/s

   
   
   

   4. 8 × 10⁵ m/s

   
   
   

---

1. View Hint View Answer Discuss in Forum

   We know that
   

   hv - φ = Kmax =	1	mv2max
   	2

   
   According to question
   

   	5hv0 - hv0	=	v22
   	2hv0 - hv0		v12

   
   v₂ = 2v₁ = 2 × 4× 10⁶ = 8 × 10⁶ m/s.

   Correct Option: C

   We know that
   

   hv - φ = Kmax =	1	mv2max
   	2

   
   According to question
   

   	5hv0 - hv0	=	v22
   	2hv0 - hv0		v12

   
   v₂ = 2v₁ = 2 × 4× 10⁶ = 8 × 10⁶ m/s.

---