MHT-CET Physics and Chemistry Full Test-6 PHY-Que-28 Solution

Q.28. Maximum kinetic energy of photon is ‘E’ when wavelength of incident radiation is ‘λ’. If wavelength of incident radiations is reduced to  \frac{\lambda}{3}  then energy of photon becomes four times. Then work function of the metal is

A. \frac{3hc}{\lambda}

B. \frac{\mathrm{hc}}{3 \lambda}

C. \frac{\mathrm{hc}}{\lambda}

D. \frac{\mathrm{hc}}{2 \lambda}

Answer: B. \frac{\mathrm{hc}}{3 \lambda}

Explanation :-

\mathrm{E} = \frac{\mathrm{hc}}{\lambda} - \phi_0

Given: \lambda = \frac{\lambda}{3} and \mathrm{E} = 4 \mathrm{E}

4 \mathrm{E} = \frac{\mathrm{hc}}{\lambda / 3} - \phi_0 \quad \ldots \text{(ii)} = \frac{3 \mathrm{hc}}{\lambda} - \phi_0 4\left(\frac{\mathrm{hc}}{\lambda} - \phi_0\right) = \frac{3 \mathrm{hc}}{\lambda} - \phi_0 \quad \ldots \text{(From (i))} \frac{4 \mathrm{hc}}{\lambda} - 4 \phi_0 = \frac{3 \mathrm{hc}}{\lambda} - \phi_0 \frac{\mathrm{hc}}{\lambda} = 3 \phi_0 \phi_0 = \frac{\mathrm{hc}}{3 \lambda}