In a photoemissive cell, with exciting wavelength `lambda`, the faster electron has speed v. If the exciting wavelength is changed to `3lambda//4`, the speed of the fastest electron will be

Options

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

Answer

D

Explanation

In photoelectric experiment, speed of fastest emitted electron is given by-
`(1)/(2)mv_(max)^(2)=(hc)/(lambda)-w`
Case-I: `(1)/(2)mv^(2)=(hc)/(lambda)-w`......(i)
Case-II: `(1)/(2)mv^(2)=(hc)/(3lambda//4)-w`
`(1)/(2)mv^(2)=(4hc)/(3lambda)-w`
From eqn.(i) & (ii)
`V'^(2)=(4)/(3)v^(2)+(w)/(3)`
Hence `v' gt v sqrt((4)/(3))`