Question

The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of $0.35V$ when the radiation $256.7nm$ is used. The work function for silver metal is

• A. $5.48eV$
• B. $2.35eV$
• C. $4.48eV$
• D. $7.35eV$

Answer 1

Answer: (C)

Energy of incident radiation (E)
$=\frac{hc}{\lambda }=\frac{6.626\times 10^{-34}\times 3\times 10^8}{256.7\times 10^{-9}}$
$=0.0774\times 10^{-17}\mathrm{J}$
$=7.74\times 10^{-19}\mathrm{J}=4.83\mathrm{eV}$
$\left[\because 1.602\times 10^{-19}\mathrm{J}=1\mathrm{eV}\right]$
The potential applied gives the kinetic energy to the electron.
$K.E$. of photoelectron $=e{V}_0$
$=1.6\times 10^{-19}\times 0.35=5.6\times 10^{-20}\mathrm{J}$
$=\frac{5.6\times 10^{-20}}{1.602\times 10^{-19}}$
$=0.349\approx 0.35\mathrm{eV}$
$\therefore$ Work function $=(4.83-0.35)\mathrm{eV}$
$=4.48\mathrm{eV}$