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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 lamda)`

B

`(hc)/(5 lamda)`

C

`(hc)/(4 lamda)`

D

`(2 hc)/(4 lamda)`

Text Solution

Verified by Experts

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