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A photon of energy 2.5 eV and wavelenght...

A photon of energy `2.5 eV` and wavelenght `lambda` falls on a metal surface and the ejected electron have velocity `'v'`. If the `lambda` of the incident light is decreased by `20%` the maximum velocity of the emitted electrons is doubled. The work function of the metal is

A

`2.6 eV`

B

`2.23 eV`

C

`2.5 eV`

D

`2.29 eV`

Text Solution

Verified by Experts

The correct Answer is:
D
`v_(1)^(2)/v_(2)^(2)=((hc)/lambda_(1)-omega)/((hc)/lambda_(2)-omega)` use`E=12400/(lambda(A^(0))`
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