Light of wavelength _ ph falls on a cathode plate inside a vacuum… - Vidyadip

MCQ

Light of wavelength $\lambda_{\text {ph }}$ falls on a cathode plate inside a vacuum tc:be 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 th: electrons passing through the anode is $\lambda_c$, which of the following statement($s$) is(are) true?

A
$\lambda_e$ decreases with increase in $\phi$ and $\lambda_{\text {ph }}$
B
$\lambda_c$ is approximately halved, if $d$ is doubled
✓
For large potential difference $(V \gg \phi / e), \lambda_e$ is approximately halved if $V$ is made four times
D
$\lambda_e$ increases at the same rate as $\lambda_{\text {ph }}$ for $\lambda_{\text {ph }} < h c / \phi$

✓

Answer

Correct option: C.

For large potential difference $(V \gg \phi / e), \lambda_e$ is approximately halved if $V$ is made four times

c
The correct option is A For large potential difference $\left(V \gg \frac{\phi}{e}\right), \lambda_e$ is approximately halved if V is made four times According to the conservation of energy principle,

$\frac{h c}{\lambda_{P h}}-\phi+e V=\frac{p_{\max }^2}{2 m}$

Now, the de-Broglie wavelength is given by,

$\lambda_e=\frac{h}{p_{\max }} \Rightarrow p_{\max }^2=\frac{h^2}{\lambda_e^2}$

$\therefore \frac{h c}{\lambda_{ Ph }}-\phi+e V=\frac{h^2}{2 m \lambda_e^2}$

When $V >> \frac{\phi}{e}$

$\Rightarrow \phi \ll e V \text { and } \frac{h c}{\lambda_{p h}}+e V=\frac{h^2}{2 m \lambda_e^2}$

$\Rightarrow \lambda_e \propto \frac{1}{\sqrt{V}}$

So, when $V$ is made four times, $\lambda_e$ is halved.

Hence, $(A)$ is the correct answer.

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