Physics - Modern Physics Question with Solution | TestHub

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by

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

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

$\frac{1}{2}m{V}^{\prime 2}=\frac{hc}{\frac{2\lambda }{3}}-\varphi =\frac{3hc}{2\lambda }-\varphi$

Substituting $\frac{hc}{\lambda }=\frac{1}{2}m{V}^2+\varphi$ into the above equation, we get

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

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

Since the work function $\varphi >0$, it follows that

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

$V^{\prime 2}>1.5V^2$

$V^{\prime }>V(1.5{)}^{\frac{1}{2}}$

Thus, the maximum speed of emitted photoelectrons will be greater than $V(1.5{)}^{\frac{1}{2}}$.