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 $\phi$ 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 $\lambda_{ ph }< hc / \phi$
• B. $\lambda_{ e }$ is approximately halved, if $d$ is doubled
• C. $\lambda_{e}$ decreases with increase in $\phi$ and $\lambda_{\text {ph }}$
• D. For large potential difference $(V \gg \phi / e), \lambda_{e}$ is approximately halved if $V$ is made four times

Answer 1

Answer: (D)

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