Question

Light of wavelength ${\lambda }_{ph}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\varphi$ and the anode is a wire mesh of conducting material kept at a distance $d$ from the cathode. A potential difference $V$ is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is ${\lambda }_e$, which of the following statements(s) is(are) true ?

• A. ${\lambda }_e$ increases at the same rate as ${\lambda }_{\text{ph }}$ for ${\lambda }_{ph}<hc/\varphi$
• B. ${\lambda }_e$ is approximately halved, if $d$ is doubled
• C. ${\lambda }_e$ decreases with increase in $\varphi$ and ${\lambda }_{\text{ph }}$
• D. For large potential difference $(V\gg \varphi /e),{\lambda }_e$ is approximately halved if $V$ is made four times

Answer 1

Answer: (D)

When electron reaches at anode Kinetic energy of electron
$k=\frac{hc}{\lambda }-\varphi +eV$
or $\frac{P^2}{2m}=\frac{hc}{\lambda }+eV-\varphi$
$\frac{h^2}{2m{\lambda }_e^2}=\frac{hc}{\lambda }+eV-\varphi$ ...(1)
if $V>\frac{\varphi }{e}$
or $eV>\varphi$
hence from equation (1)
$\frac{h^2}{2m{\lambda }_e^2}=\frac{hc}{\lambda }+eV$
$\frac{h^2}{2m{\lambda }_e^2}=eV$
$\left(\because eV>\varphi \text{ or }eV>\frac{hc}{\lambda }\right)$
i.e. ${\lambda }_e\propto \frac{1}{\sqrt{V}}$