Physics - Modern Physics Question with Solution | TestHub

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K=\frac{hc}{\lambda }-\Phi$, where $\Phi$ is the work function of the material.

For the first case, the maximum kinetic energy is:

$K_1=\frac{hc}{\lambda }-\Phi$

For the second case, the wavelength is $\lambda /3$, so the maximum kinetic energy is:

$K_2=\frac{hc}{\lambda /3}-\Phi =\frac{3hc}{\lambda }-\Phi$

It is given that $K_2=4K_1$. Substituting the expressions for $K_1$ and $K_2$, we get:

$\frac{3hc}{\lambda }-\Phi =4\left(\frac{hc}{\lambda }-\Phi \right)$

$\frac{3hc}{\lambda }-\Phi =\frac{4hc}{\lambda }-4\Phi$

$4\Phi -\Phi =\frac{4hc}{\lambda }-\frac{3hc}{\lambda }$

$3\Phi =\frac{hc}{\lambda }$

$\Phi =\frac{hc}{3\lambda }$

Answer: $\frac{\mathrm{hc}}{3\lambda }$