Photo electrons are emitted with a maximum speed of `7xx10^(5)ms^(-1)` from a surface when light of frequency `8xx10^(14)Hz` is incident on it, the threshold frequency for this surface is :

Options

A. `2.32xx10^(14)Hz`
B. `4.64xx10^(14)Hz`
C. `4.64xx10^(20)Hz`
D. `6.64Hz`

Answer

B

Explanation

`h(v-v_(0))=(1)/(2) mv_("max")^(2) or v_(0)=v-(mv_("max")^(2))/(2h)`
`=8xx10^(14)-(9.1xx10^(-31)xx49xx10^(10))/(2xx6.63xx10^(-34))`
`=8xx10^(14)3.63xx10^(14)=4.64xx10^(14)Hz`