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

Light of wavelength `4000 Å` falls on a photosensitive metal anda 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 ^(-10) 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) = 2eV`
Hence by using `E = W _(0) + K _(max)`
`W _(0)= 1 ,09 eV ~~1. 1 eV`
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