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A light whose frequency is equal to `6xx10^(14)` Hz is incident on a metal whose work function is 2 eV.
`[h=6.63 xx10^(-34) J s, 1eV = 1.6 " The" xx10^(-19)].`
maximum energy of the electrons emitted will be

A

2.49

B

4.49 eV

C

0.49 eV

D

5.49 eV

Text Solution

Verified by Experts

The correct Answer is:
C
`k_(max) = hv -phi_0`
`=(6.63 xx10^(-34)xx6 xx10^(14))/(1.6 xx10^(-19))-2 =2.49 -2 =0.49 eV`
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