11. The number of photo electrons emitted for light of a frequency v (higher than the threshold frequency v₀) is proportional to:

1. 1. Threshold frequency (v₀)

   
   
   

   2. ​Intensity of light

   
   
   

   3. Frequency of light (v)

   
   
   

   4. v – v₀

   
   
   

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1. View Hint View Answer Discuss in Forum

   The number of photoelectrons emitted is proportional to the intensity of incident light. Saturation current ∝ intensity.

   Correct Option: B

   The number of photoelectrons emitted is proportional to the intensity of incident light. Saturation current ∝ intensity.

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12. A source S₁ is producing, 10¹⁵ photons per second of wavelength 5000 Å. Another source S₁ is producing 1.02×10¹⁵ photons per second of wavelength 5100 Å Then,
    (power of S₂) (power of S₁) is equal to :​​

1. 1. 1.00

   
   
   

   2. ​1.02 ​

   
   
   

   3. 1.04 ​

   
   
   

   4. 0.98

   
   
   

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1. View Hint View Answer Discuss in Forum

   Energy emitted/sec by
   

   S1, P1 = n1	hc
   	λ1

   
   Energy emitted/sec by
   

   S2, P2 = n2	hc
   	λ2

   
   

   ∴	P2	=	n2	.	λ1
   	P1		n1		λ2

   
   

   =	1.02 × 1015	.	5000	= 1.0
   	1015		5100

   Correct Option: A

   Energy emitted/sec by
   

   S1, P1 = n1	hc
   	λ1

   
   Energy emitted/sec by
   

   S2, P2 = n2	hc
   	λ2

   
   

   ∴	P2	=	n2	.	λ1
   	P1		n1		λ2

   
   

   =	1.02 × 1015	.	5000	= 1.0
   	1015		5100

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13. The potential difference that must be applied to stop the fastest photo electrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:

1. 1. 2.4 V ​

   
   
   

   2. – 1.2 V

   
   
   

   3. – 2.4 V

   
   
   

   4. 1.2 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   Kmax =	hc	- W =	hc	- 5.01
   	λ		λ

   
   

   =	12375	- 5.01
   	λ(in Å)

   
   

   =	12375	- 5.01
   	2000

   
   = 6.1875 – 5.01 = 1.17775 ≃ ​1.2 V

   Correct Option: D

   Kmax =	hc	- W =	hc	- 5.01
   	λ		λ

   
   

   =	12375	- 5.01
   	λ(in Å)

   
   

   =	12375	- 5.01
   	2000

   
   = 6.1875 – 5.01 = 1.17775 ≃ ​1.2 V

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14. The threshold frequency for a photosensitive metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly

1. 1. 2 V

   
   
   

   2. 3 V ​

   
   
   

   3. 5 V

   
   
   

   4. 1 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   K.E. = hν – hν_th = eV₀   (V₀ = cut off voltage)
   

   ⇒ V0 =	h	(8.2 × 1014 - 3.3× 1014)
   	e

   
   

   =	6.6 × 10-34 × 4.9 × 1014	≈ 2V.
   	1.6 × 10-19

   Correct Option: A

   K.E. = hν – hν_th = eV₀   (V₀ = cut off voltage)
   

   ⇒ V0 =	h	(8.2 × 1014 - 3.3× 1014)
   	e

   
   

   =	6.6 × 10-34 × 4.9 × 1014	≈ 2V.
   	1.6 × 10-19

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15. The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three different radiations. Which one of the following is a correct statement?
    

1. 1. Curves (1) and (2) represent incident radiations of same frequency but of different intensities.

   
   
   

   2. Curves (2) and (3) represent incident radiations of different frequencies and different intensities.

   
   
   

   3. Curves (2) and (3) represent incident radiations of same frequency having same intensity.

   
   
   

   4. Curves (1) and (2) represent incident radiations of different frequencies and different intensities.

   
   
   

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1. View Hint View Answer Discuss in Forum

   Retarding potential depends depends on the frequency of incident radiation but is independent of intensity.

   Correct Option: A

   Retarding potential depends depends on the frequency of incident radiation but is independent of intensity.

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