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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 xx 10^(-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) = 3.09 eV`
Stopping potential is `- 2V` so `K_(max) = 2 eV`
Hence by using `E = W_(0) + K_(max) rArr W_(0) = 1.09 ~~ 1.1` eV
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Knowledge Check

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