Radiation of frequency `v=9v_(0)` is incident on a metal surface, having threshold frequency `v_(0)`. The maximum velocity of the ejected photoelectrons is `6xx10^(6)m//s`. What will be the maximum velocity of the ejected photoelectrons , if v is reduced to `3v_(0)` ?

Options

A. `2xx10^(6)` m/s
B. `3xx10^(6)` m/s
C. `1.5xx10^(6)` m/s
D. `4xx10^(6)` m/s

Answer

B

Explanation

`1/2mv_(1)^(2)=h(v-v_(0))=h(9v_(0)-v_(0))=8v_(0)h`
and `1/2mv_(2)^(2)=h(3v_(0)-v_(0))=2v_(0)h`
`:.(v_(2)^(2))/(v_(1)^(2))=1/4:.v_(2)=(v_(1))/(2)=3xx10^(6)m//s`