Radiation of wavelength `lambda` is incident on a photocell.The fastest emitted electron has
speed v.If wavelength is changed to `(3lambda)/(4)`
the speed of the fastest emitted electron will be

Options

A. `vsqrt((3)/(4))`
B. `vsqrt((4)/(3))`
C. less than `v sqrt((3)/(4))`
D. greater than `vsqrt((4)/(3))`

Answer

D

Explanation

`(hc)/(lambda) = phi_(0) + 1/2mv^(2) , (4hc)/(3lambda) = phi_(0) + 1/2 mv_(1)^(2)`
`rArr 0 = 4/3phi_(0) - phi_(0) + 4/3(1/2mv^(2)) = 1/2mv_(1)^(2)`
`rArr 1/2mv_(1)(2) = 4/3 (1/2mv^(2)) + (phi_(0))/(3) rArr v_(1) gt vsqrt((4)/(3))`