The threshold frequency for certain metal is `v_0`. When light of frequency `2v_0` is incident on it, the maximum velocity of photoelectrons is `4xx10^(6) ms^(-1)`. If the frequency of incident radiation is increaed to `5v_0`, then the maximum velocity of photoelectrons will be

Options

A. `(4//5)xx10^(6)`
B. `2xx10^(6)`
C. `4xx10^(6)`
D. `8xx10^(6)`

Answer

d

Explanation

As `1/2mv_(max)^(2)=hv-hv_(0)`
`:. 1/2mxx(4xx10^(6))^(2)=h2v_(0)-hv_(0)=hv_(0)`.......(i)
and `1/2mv'_(max)^(2)=h(2v_(0)-v_(0))-hv_(0)=4hv_(0)`........(ii)
Dividing (i) by (i), we have
`(v'_(max)^(2))/((4xx10^(6))^(2))=4` or `v'_(max)=2xx4xx10^(6)`
`=8xx10^(6)ms^(-1)`