71. As the intensity of incident light increases

1. 1. photoelectric current increases

   
   
   

   2. K. E. of emitted photo electrons increases

   
   
   

   3. photoelectric current decreases

   
   
   

   4. K.E. of emitted photo electrons decreases

   
   
   

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1. View Hint View Answer Discuss in Forum

   K.E. of electrons emitted depends upon the frequency of incident rays rather than the intensity. While number of photo electrons emitted depends upon intensity of radiation.

   Correct Option: A

   K.E. of electrons emitted depends upon the frequency of incident rays rather than the intensity. While number of photo electrons emitted depends upon intensity of radiation.

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72. The photoelectric work function for a metal surface is 4.125 eV. The cut off wavelength for this surface is

1. 1. 4125 Å

   
   
   

   2. 3000 Å

   
   
   

   3. 6000 Å

   
   
   

   4. 2062.5 Å

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let λ₀ be cut off wavelength.
   Work function
   

   =	hc	= 4.125 × 1.6 × 10-19
   	λ0

   
   

   λ0 =	6.626 × 10-34 × 3 × 108	= 3000 Å
   	4.125 × 1.6 × 10-19

   Correct Option: B

   Let λ₀ be cut off wavelength.
   Work function
   

   =	hc	= 4.125 × 1.6 × 10-19
   	λ0

   
   

   λ0 =	6.626 × 10-34 × 3 × 108	= 3000 Å
   	4.125 × 1.6 × 10-19

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73. In a photo-emissive cell, with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4 , the speed of the fastest emitted electron will be

1. 1. (3/4)^1/2. v

   
   
   

   2. (4/3)^1/2. v

   
   
   

   3. less than (4/3)^1/2. v

   
   
   

   4. greater than (4/3)^1/2. v

   
   
   

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1. View Hint View Answer Discuss in Forum

   	1	mv2 =	hc	- W0
   	2		λ

   
   

   or	hc	=	1	mv2 + W0
   	λ		2

   
   

   and	1	mv21 =	hc	- W0
   	2		(3λ/4)

   
   

   	1			1	mv2 + W0		- W0
   	4			2

   
   So, v₁ is greater than
   

   v1 =		4		1/2
   		3

   Correct Option: D

   	1	mv2 =	hc	- W0
   	2		λ

   
   

   or	hc	=	1	mv2 + W0
   	λ		2

   
   

   and	1	mv21 =	hc	- W0
   	2		(3λ/4)

   
   

   	1			1	mv2 + W0		- W0
   	4			2

   
   So, v₁ is greater than
   

   v1 =		4		1/2
   		3

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74. The 21 cm radio wave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly

1. 1. 10^–17 J

   
   
   

   2. 1 J

   
   
   

   3. 7 × 10^–8 J

   
   
   

   4. 10^–24 J

   
   
   

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1. View Hint View Answer Discuss in Forum

   E = hv =	hc	=	6.6 × 10-34 × 3 × 108
   	λ		0.21

   
   

   =	6.6	× 10-24 = 0.94 × 10-24 J
   	7

   Correct Option: D

   E = hv =	hc	=	6.6 × 10-34 × 3 × 108
   	λ		0.21

   
   

   =	6.6	× 10-24 = 0.94 × 10-24 J
   	7

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75. Light of wavelength 5000 Å falls on a sensitive plate with photo-electric work function of 1.9 eV. The kinetic energy of the photo-electrons emitted will be

1. 1. 0.58 eV

   
   
   

   2. 2.48 eV

   
   
   

   3. 1.24 eV

   
   
   

   4. 1.16 eV

   
   
   

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1. View Hint View Answer Discuss in Forum

   K.E. = hν – hν₀
   

   =	hc	– hν0 =	12375	- 1.9
   	λ		5000

   
   = 2.48 – 1.9 = 0.58 eV

   Correct Option: A

   K.E. = hν – hν₀
   

   =	hc	– hν0 =	12375	- 1.9
   	λ		5000

   
   = 2.48 – 1.9 = 0.58 eV

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