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When a metallic surface is illuminated w...

When a metallic surface is illuminated with monochromatic light of wavelength `lambda`, the stopping potential is `5V_0`. When the same surface is iluminated with light of wavelength `3lambda`, the stopping potential is `V_0`. The work function of the metallic surface is

A

`(hc)/(6 lambda)`

B

`(hc)/(5lambda)`

C

`(hc)/(4lambda)`

D

`(2hc)/(4lambda)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(hc)/(lambda) =5 eV_(0)+phi`
`(hc)/(3 lambda) =eV_(0) +phirArr (2hc)/(3lambda) = 4eV_(0) rArr phi=(hc)/(6 lambda) `
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