Question

A radiation of wave length 2500 $A^0$ is incident on a metal plate whose work function is 3.5 eV. Then the potential required to stop the fastest photo electrons emitted by the surface is :

($h=6.63\times {10}^{-34}$Js and $c=3\times {10}^8$ m/s)

• A. 2.94 V
• B. 1.47 V
• C. 3.5 V
• D. 4.9 V

Answer 1

The correct option is B 1.47 V

Energy of photon $E=h\nu =\frac{hc}{\lambda }$
Given:
$\lambda =2500\times {10}^{-10}m$

$E=\frac{6.63\times 3\times {10}^{-26}}{2500\times {10}^{-10}}J$

$E=\frac{0.007956\times {10}^{-16}}{1.6\times {10}^{-19}}eV=4.97eV$

Work function of metal is $3.5eV$.

So maximum kinetic energy of the photoelectron after emission $K=E-W=(4.97-3.5)eV=1.47eV$
So the voltage required to stop the fastest photoelectron is $1.47V$.

So, the answer is option (B).