Threshold frequency for photosensitive material is `v_(0)`. If photons of frequency `2v_(0)` fall on this surface, the electrons come out with a maximum speed of `4xx10^(6)ms^(-1)`. When photons of frequency `5v_(0)` fall on the same surface, the maximum speed of the ejected electrons will be

Options

A. `2xx10^(7)ms^(-1)`
B. `2xx10^(6)ms^(-1)`
C. `8xx10^(6)ms^(1)`
D. `8xx10^(5)ms^(-1)`

Answer

C

Explanation

As per relation `h(v-v_(0))=(1)/(2)mv_(max)^(2)`, we have
`((v_(1))_(max))/((v_(2))_(max))=sqrt((v_(1)-v_(0))/(v_(2)-v_(0)))=sqrt((2v_(0)-v_(0))/(5v_(0)-v_(0)))=1/2 implies (v_(2))_(max)=2(v_(1))_(max)=2xx4xx10^(6)=8xx10^(6)ms^(-1)`.