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Photons of energy 2.0eV fall on a metal ...

Photons of energy `2.0eV` fall on a metal plate and release photoelectrons with a maximum velocity `V`. By decreasing `lambda` and `25%` the maximum velocity of photoelectrons is doubled. The work function of the metal of the material plate in `eV` is nearly

A

`2.22`

B

`1.985`

C

`2.35`

D

`1.80`

Text Solution

Verified by Experts

The correct Answer is:
D
`1/2mv_(max)^(2)=hv-w` & `hv=(hc)/lambda`
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