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					 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
 
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- (3/4)1/2. v
 - (4/3)1/2. v
 - less than (4/3)1/2. v
 - greater than (4/3)1/2. v
 
 
Correct Option: D
| mv2 = | - W0 | |||
| 2 | λ | 
| or | = | mv2 + W0 | ||
| λ | 2 | 
| and | mv21 = | - W0 | ||
| 2 | (3λ/4) | 
![]()  | mv2 + W0 | ![]()  | - W0 | ||||
| 4 | 2 | 
So, v1 is greater than
| v1 = | ![]()  | ![]()  | 1/2 | 3 | 

