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Ultraviolet radiations of 6.2 eV fall on...

Ultraviolet radiations of 6.2 eV fall on an aluminium surface (work function 4.2 eV ) . The kinetic energy (in joule ) of the fastest electron emitted is approximately

A

`3.2 xx10^(-21)`

B

`3.2 xx10^(-19)`

C

`3.2 xx10^(-17)`

D

`3.2 xx10^(-15)`

Text Solution

Verified by Experts

The correct Answer is:
B
`K_("max") = hv - phi_0 implies K_("max")= (6.2 -4.2)eV =2eV`
`=2xx1.6 xx10^(-19)J = 3.2 xx10^(-19) J`
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