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Ultraviolet light of 6.2eV falls on Caes...

Ultraviolet light of `6.2eV` falls on Caesium surface (work function `= 1.2eV)`. The kinetic energy (in electron volts) of the fastest electron emitted is approximately

A

`5eV`

B

`4 eV`

C

`3 eV`

D

` 2 eV`

Text Solution

Verified by Experts

The correct Answer is:
A
K.E = E - W
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