Physics - Modern Physics Question with Solution | TestHub

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

For the first case with wavelength $\lambda$:
$K_1=\frac{hc}{\lambda }-\varphi$

For the second case with wavelength $\frac{\lambda }{2}$:
$K_2=\frac{hc}{\left(\frac{\lambda }{2}\right)}-\varphi =\frac{2hc}{\lambda }-\varphi$

Given that $K_1=\frac{1}{4}{K}_2$, we have $4K_1=K_2$.

Substituting the expressions for $K_1$ and $K_2$:
$4\left(\frac{hc}{\lambda }-\varphi \right)=\frac{2hc}{\lambda }-\varphi$

$\frac{4hc}{\lambda }-4\varphi =\frac{2hc}{\lambda }-\varphi$

$\frac{4hc}{\lambda }-\frac{2hc}{\lambda }=4\varphi -\varphi$

$\frac{2hc}{\lambda }=3\varphi$

$\varphi =\frac{2hc}{3\lambda }$

Answer: $\frac{2\mathrm{hc}}{3\lambda }$