Question

The ejection of a photoelectron from a metal in the photoelectric effect experiment can be stopped by applying $0.25\text{eV}$. used. The work function of the metal is 3.5 eV.
The wavelength of the incident radiation is:
(Take $h=6.6\times {10}^{-34}\text{J s}$)

• A. $3.3\times {10}^{-7}\text{m}$
• B. $3.0\times {10}^{-7}\text{m}$
• C. $3.3\times {10}^{-6}\text{m}$
• D. $3.0\times {10}^{-6}\text{m}$

Answer 1

The correct option is A $3.3\times {10}^{-7}\text{m}$

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

The potential applied gives kinetic energy to the electron.
Hence, kinetic energy of the electron = $0.25\text{eV}$.
$\therefore \text{Energy of incident radiation}=3.5\text{eV}+0.25\text{eV}=3.75\text{eV}$
= $3.75\times 1.6\times {10}^{-19}J=6\times {10}^{-19}\text{J}$

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

$6\times {10}^{-19}=\frac{(6.6\times {10}^{-34}\text{J s})(3.0\times {10}^8{\text{m s}}^{-1})}{\lambda }$

$\lambda =3.3\times {10}^{-7}\text{m}$