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The work function of cesium is 2.27 eV. ...

The work function of cesium is 2.27 eV. The cut-off voltage wich stops the emission of electrons from a cesium cathode irradiated with light of 600 nm wavelength is

A

0.5 V

B

`-0.2V`

C

`- 0.5V`

D

`0.2V`

Text Solution

Verified by Experts

Here, `phi` =2.27 eV, `lambda` = 600 nm
V ?
According to Einstein .s photoelectric equation
`KE_(max) = (hc)/(lambda)-phie = eV`
Energy of incident radiation,
`(hc)/(lambda) = (1242e Vnm)/(600 nm ) = 2.07eV lt phi`
Hence no emission of electrons will take place with light of `lambda = 600` nm.
None of the given options is correct.
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