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

Absorbed energy=Threshold energy+Kinetic energy of photoelectrons
`hv=hv_(0)+KE`
`hv_(0)=hv-KE`
`6.626xx10^(-34)xxv_(0)=6.626xx10^(-34)xx2xx10^(14)-6.63xx10^(-20)`
`v_(0)=(1.3252xx10^(-19)-6.63xx10^(-20))/(6.626xx10^(-34))`
`v_(0)=9.99xx10^(13)=10^(14)s^(-1)`.