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

Options

A. `4xx10^(6)m" "s^(-1)`
B. `6xx10^(6)m" "s^(-1)`
C. `8xx10^(6)m" "s^(-1)`
D. `1xx10^(6)m" "s^(-1)`

Answer

A

Explanation

(a) : According to Einstein's photoelectric equation
`hv=hv_(0)+(1)/(2)mv_(max)^(2) or (1)/(2)mv_(max)^(2)=hv-hv_(0)`
According to the given problem
`(1)/(2)m(8xx10^(6))^(2)=h(5v_(0)-v_(0))" ". . .(i)`
`(1)/(2)mv_(max)^(2)=h(2v_(0)-v_(0))" ". . . (ii)`
Dividing equation (i) by (ii), we get
`((8xx10^(6))^(2))/(v_(max)^(2))=(4v_(0))/(v_(0))rArrv_(max)^(2)=((8xx10^(6))^(2))/(4)`
`v_(max)=((8xx10)^(6))/(2)=4xx10^(6)"m s"^(-1)`