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In a photoemissive cell, with exciting w...

In a photoemissive cell, with exciting wavelength `lambda`, the faster electron has speed v. If the exciting wavelength is changed to `3lambda//4`, the speed of the fastest electron will be

A

`vsqrt(3/4)`

B

`vsqrt(4/3)`

C

less than `vsqrt(3/4)`

D

greater than `vsqrt(4/3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`1/2mv^(2)=(hv)/lambda-phi,1/2mv'^(2)=(hv)/((3lambda//4))-f=(4hc)/(3lambda)-phi` clearly `v' gt sqrt(4/3)v`
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