Question

For a certain metal v is the five times of `v_(0)` and the maximum velocity of coming out photons is `8 xx 10^(6)m//s`. If `v = 2v_(0)`, then maximum velocity of photoelectrons will be
A. `4 xx 10^(6)m//s`
B. `6xx10^(6)m//s`
C. `2xx10^(6)m//s`
D. `1xx10^(6)m//s`

Answer 1

Correct Answer - A
`(1)/(2)m(8xx10^(6))^(2)=h(5v_(0)-v_(0))" "...(i)`
`" and "(1)/(2)mv^(2)=h(2v_(0)-v_(0))" "...(ii)`
Divding Eq. (ii) by Eq. (i), we get
`((8xx10^(6))^(2))/(v^(2))=(4v_(0))/(v_(0))impliesv_(2)=((8xx10^(6))^(2))/(4)`
`v=(8xx10^(6))/(2)impliesv=4xx10^(6 )m//s`