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 statement(s) is(are) true?

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

Answer 1

Answer: (D)

${\lambda }_e=$ minimum be broglie wavelength.

$\frac{hc}{{\lambda }_{ph}}=\frac{hc}{{\lambda }_e}+\varphi +eV$

$\Rightarrow$ $hc\left(\frac{-1}{{\lambda }_{ph}^2}d{\lambda }_{ph}\right)=hc\left(\frac{-1}{{\lambda }_e^2}d{\lambda }_e\right)$

$\Rightarrow$ $\frac{d{\lambda }_{ph}}{d{\lambda }_e}=\frac{{\lambda }_{ph}^2}{{\lambda }_e^2}$

This denies option A, B and C.

$V>>\frac{\varphi }{e}$

Energy of electron,

$E=\frac{P^2}{2m}$

${\lambda }_e=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$$=\frac{h}{\sqrt{2meV}}$

This supports option D.