Physics - Modern Physics Question with Solution | TestHub

According to Einstein's photoelectric equation:

$K_{\max }=\frac{hc}{\lambda }-\Phi$

Since the work function $\Phi$ is negligible, $\Phi =0$.

$\frac{1}{2}m{V}^2=\frac{hc}{\lambda }$

When the incident wavelength is changed to ${\lambda }^{\prime }=\frac{2\lambda }{3}$, the new maximum kinetic energy is:

$\frac{1}{2}m(V^{\prime }{)}^2=\frac{hc}{{\lambda }^{\prime }}=\frac{hc}{\frac{2\lambda }{3}}$

$\frac{1}{2}m(V^{\prime }{)}^2=\frac{3}{2}\left(\frac{hc}{\lambda }\right)$

Substituting the value of $\frac{hc}{\lambda }$:

$\frac{1}{2}m(V^{\prime }{)}^2=\frac{3}{2}\left(\frac{1}{2}m{V}^2\right)$

$(V^{\prime }{)}^2=\frac{3}{2}{V}^2$

$V^{\prime }=\sqrt{\frac{3}{2}}V$

Answer: $\sqrt{\frac{3}{2}}V$