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A photon of energy 8 eV is incident on m...

A photon of energy 8 eV is incident on metal surface of threshold frequency `1.6 xx10^(15)` Hz, The kinetic energy of the photoelectrons emitted (in eV): (Take `h = 6 xx 10 ^(-34) J-s)`

A

`1.6`

B

`6`

C

`2`

D

`1.2`

Text Solution

Verified by Experts

The correct Answer is:
C
`KE = hv - hv_(0)`
`= 8 eV - (( 6 xx 10 ^(-34) xx 1. 6 xx 10 ^(15))/( 1.6 xx 10 ^(-19))eV)`
`= 8 - 6 = 2 eV`
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