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Light of wavelength 4000 Å falls on a ph...

Light of wavelength 4000 Å falls on a photosensitive metal and a negative 2 V potential stops the emitted electrons. The work function of the material (in eV) is approximately.
`(h=6.6 xx10^(-34)Js, e = 1.6 xx 10^(-19)C, c= 3 xx 10^8 ms^(-1))`

A

`1.1`

B

`2.0`

C

`2.2`

D

`3.1`

Text Solution

Verified by Experts

The correct Answer is:
A
Energy of incident light `E(eV)= (12375)/(4000)=3.09eV`
Stopping potential is `-2V " so " K_("max")=2eV`
Hence by using `E=W_0+K_("max")`,
`W_0 =1.09eV ~~ 1.leV`
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