When light of wavelength $\lambda$ incidents on a photosensitive material, photoelectrons are emitted. If…
When light of wavelength $\lambda$ incidents on a photosensitive material, photoelectrons are emitted. If the wavelength of the incident light is reduced by $50 \%$, the maximum kinetic energy of the emitted photoelectrons becomes 3 times the initial maximum kinetic energy. The work function of the material is
( $\mathrm{h}$ - Planck's constant, $\mathrm{c}-$ Speed of light in vacuum)
$\frac{\mathrm{hc}}{\lambda}$
$\frac{\mathrm{hc}}{2 \lambda}$
$\frac{2 \mathrm{hc}}{\lambda}$
$\frac{\mathrm{hc}}{3 \lambda}$
Solution
From the photoelectric equation
$\frac{\mathrm{hc}}{\lambda}=\mathrm{E}+\phi...(i)$
When $\lambda^{\prime}=0.5 \lambda=\frac{\lambda}{2}$
Maximum Kinetic Energy, $\mathrm{E}^{\prime}=3 \mathrm{E}$
$\frac{2 \mathrm{hc}}{\lambda}=3 \mathrm{E}+\phi... (ii)$
Solving equation (2) by (1),
we have work function, $\phi=\frac{\mathrm{hc}}{2 \lambda}$