56. Ultraviolet radiations of 6.2 eV falls on an aluminium surface. K.E. of fastest electron emitted is (work function = 4.2 eV)

1. 1. 3.2 × 10^–21 J ​

   
   
   

   2. 3.2 × 10^–19 J ​

   
   
   

   3. ​7 × 10^–25 J

   
   
   

   4. 9 × 10^–32 J

   
   
   

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1. View Hint View Answer Discuss in Forum

   K.E. of fastest electron
   = E – W₀ = 6.2 – 4.2 = 2.0 eV ​
   = 2 × 1.6 × 10^–19 = 3.2 × 10^–19 J

   Correct Option: B

   K.E. of fastest electron
   = E – W₀ = 6.2 – 4.2 = 2.0 eV ​
   = 2 × 1.6 × 10^–19 = 3.2 × 10^–19 J

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57. The energy of a photon of wavelength λ is

1. 1. ​hc λ

   
   
   

   2. 	hc
      	λ

   
   
   

   3. 	λ
      	hc

   
   
   

   4. 	λh
      	c

   
   
   

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1. View Hint View Answer Discuss in Forum

   Energy of a photon
   

   E = hν =	hc
   	λ

   Correct Option: B

   Energy of a photon
   

   E = hν =	hc
   	λ

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58. According to Einstein’s photoelectric equation, the graph between the kinetic energy of photo electrons ejected and the frequency of incident radiation is

1. 1. 

   
   
   

   2. 

   
   
   

   3. 

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   K.E. = hν – φ​
   ∴ graph of K.E. versus ν is a straight line with +ve slope h and K.E. > 0 for ν such that hν > φ​.

   Correct Option: A

   K.E. = hν – φ​
   ∴ graph of K.E. versus ν is a straight line with +ve slope h and K.E. > 0 for ν such that hν > φ​.

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59. The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of incident radiation for which the stopping potential is 5 V lies in the:​​

1. 1. Ultraviolet region

   
   
   

   2. Visible region

   
   
   

   3. Infrared region

   
   
   

   4. X-ray region

   
   
   

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1. View Hint View Answer Discuss in Forum

   Work function, φ₀ = 6.2eV.
   Stopping potential V₀ = 5V. ​
   As, eV₀ = hv –  φ₀
   

   or eV0 + φ0 =	hc
   	λ

   
   

   or λ =	hc
   	eV0 + φ0

   
   

   λ =	6.62 × 10-34 Js × 3 × 108 ms-1
   	1.6 × 10-19 × 5J + 6.2 × 1.6 × 10-19J

   
   

   λ =	6.62 × 10-34 × 3 × 108	m
   	1.6 × 10-19 (5 + 6.2)

   
   

   =	19.86 × 10-26	1.10 × 10-7 m
   	17.92 × 10-19

   
   Thus the wavelength of the incident radiation lies in the ultraviolet region.

   Correct Option: A

   Work function, φ₀ = 6.2eV.
   Stopping potential V₀ = 5V. ​
   As, eV₀ = hv –  φ₀
   

   or eV0 + φ0 =	hc
   	λ

   
   

   or λ =	hc
   	eV0 + φ0

   
   

   λ =	6.62 × 10-34 Js × 3 × 108 ms-1
   	1.6 × 10-19 × 5J + 6.2 × 1.6 × 10-19J

   
   

   λ =	6.62 × 10-34 × 3 × 108	m
   	1.6 × 10-19 (5 + 6.2)

   
   

   =	19.86 × 10-26	1.10 × 10-7 m
   	17.92 × 10-19

   
   Thus the wavelength of the incident radiation lies in the ultraviolet region.

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60. Monochromatic light of frequency 6.0 × 10¹⁴ Hz is produced by a laser. The power emitted is 2 × 10–3 ω. The number of photons emitted, on the average, by the sources per second is

1. 1. 5 × l 0¹⁶

   
   
   

   2. 5 × 10¹⁷

   
   
   

   3. 5 × 10¹⁴

   
   
   

   4. 5 × 10¹⁵

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since p = nhν
   

   ⇒ n =	p	=	2 × 10-3	= 5 × 1015
   	hv		6.6 × 10-34 × 6 × 1014

   Correct Option: D

   Since p = nhν
   

   ⇒ n =	p	=	2 × 10-3	= 5 × 1015
   	hv		6.6 × 10-34 × 6 × 1014

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