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, is true?

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

Answer 1

The correct option is A For large potential difference $(V>>\frac{\varphi }{e}),{\lambda }_e$ is approximately halved if $V$ is made four times

According to the conservation of energy principle,

$\frac{hc}{{\lambda }_{Ph}}-\varphi +eV=\frac{p_{max}^2}{2m}$

Now, the de-Broglie wavelength is given by,
${\lambda }_e=\frac{h}{p_{max}}\Rightarrow p_{max}^2=\frac{h^2}{{\lambda }_e^2}$

$\therefore \frac{hc}{{\lambda }_{Ph}}-\varphi +eV=\frac{h^2}{2m{\lambda }_e^2}$

When $V>>\frac{\varphi }{e}$
$\Rightarrow \varphi <<eV\text{and}\frac{hc}{{\lambda }_{Ph}}+eV=\frac{h^2}{2m{\lambda }_e^2}$
$\Rightarrow {\lambda }_e\propto \frac{1}{\sqrt{V}}$
So, when $V$ is made four times, ${\lambda }_e$ is halved.

Hence, $\left(A\right)$ is the correct answer.