Calculate the velocity of the photoelectrons emitted when a light of frequency `3xx 10^(12) `Hz is incident on a metal surface of threshold frequency equal to `2xx10^(12)` Hz.

Explanation

Given, `v=3xx10^(12)`Hz, `v_(0)=2xx10^(12)` Hz.
By photoelectric equation, `hv=hv_(0)+(KE)_(max)`
`therefore (KE)_(max)=h(v-v_(0))`
`=6.626xx10^(-34)xx10^(12)(3-2)=6.626xx10^(-22)`J.
But, `K.E. = 1/2 mv^(2)`
Hence, `V^(2)=(2 KE)/m=(2xx6.626xx10^(-22))/(9.1xx10^(-31))=14.56xx10^(8)`
`therefore " " v=sqrt(14.56xx10^(8))=3.816xx10^(4)ms^(-1)`
Maximum velocity of electron `=3.8xx10^(4)ms^(-1)`.