Question

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

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

Answer 1

Answer: (C)

Energy of incident radiation (E)
$=\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{256.7 \times 10^{-9}}$
$=0.0774 \times 10^{-17} \mathrm{~J}$
$=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 $=e V_{0}$
$=1.6 \times 10^{-19} \times 0.35=5.6 \times 10^{-20} \mathrm{~J}$
$=\frac{5.6 \times 10^{-20}}{1.602 \times 10^{-19}}$
$=0.349 \approx 0.35 \mathrm{eV}$
$\therefore \quad$ Work function $=(4.83-0.35) \mathrm{eV}$
$=4.48 \mathrm{eV}$