The maximum kinetic energy of photoelectrons ejected from a metal, when it is irradiated with radiation of frequency `2xx10^(14)s^(-1)` is `6.63xx10^(-20)`J. the threshold frequency of the metal is:

Options

A. `2xx10^(14)s^(-1)`
B. `3xx10^(14)s^(-1)`
C. `2xx10^(-14)s^(-1)`
D. `1xx10^(14)s^(-1)`

Answer

D

Explanation

Maximum kinetic energy = Absorbed energy - Threshold energy
`6.63xx10^(-20)J=hv-hv_(0)`
`6.63xx10^(-20)J=h(2xx10^(14)-v_(0))`
`2xx10^(14)-v_(0)=(6.63xx10^(-20))/(6.62xx10^(-34))`
`2xx10^(14)-v_(0)=10^(14)`
`v_(0)=1xx10^(14)s^(-1)`.