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The surface of a metal illuminated with the light of `400 nm`. The kinetic energy of the ejected photoelectrons was found to be `1.68 eV`. The work function of the metal is : `(hc=1240 eV.nm)`

A

`1.41 eV`

B

`1.51 eV`

C

`1.68 eV`

D

`3.09 eV`

Text Solution

Verified by Experts

The correct Answer is:
A
`E_(lambda)=(1240)/(400)eV= 3.1 eV`
`E_(lambda)-k(3.10-1.68)eV`
`=1.42 eV`
`Q lt 1.42 eV`
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