Question

The ejection of a photoelectron from a metal in the photoelectric effect experiment can by stopped by applying $0.35\text{eV}$ when a radiation of $256.7\text{nm}$ is used. The work function of the metal is:

• A. $4\text{eV}$
• B. $5.5\text{eV}$
• C. $4.5\text{eV}$
• D. $5\text{eV}$

Answer 1

The correct option is C $4.5\text{eV}$

$\text{Energy of the incident radiation}\text{= Work function + Kinetic energy of photoelectron}$

Energy of incident radiation, $\left(E\right)=h\nu =\frac{hc}{\lambda }$

$=\frac{(6.626\times {10}^{-34}\text{J s})(3.0\times {10}^8{\text{m s}}^{-1})}{256.7\times {10}^{-9}\text{m}}$

$=7.74\times {10}^{-19}\text{J}$
$=4.83\text{eV}(\because 1\text{eV}=1.602\times {10}^{-19}\text{J})$

The potential applied gives kinetic energy to the electron.
Hence, kinetic energy of the electron = $0.35\text{eV}$.
$\therefore \text{Work function}=4.83\text{eV}-0.35\text{eV}=4.48\text{eV}$