Modern Physics Ques 181
- 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?
(2016 Adv.)
(a) $\lambda _e$ increases at the same rate as $\lambda _{ph}$ for $\lambda _{ph}<h c / \phi$
(b) $\lambda _e$ is approximately halved, if $d$ is doubled
(c) $\lambda _e$ decreases with increase in $\phi$ and $\lambda _{ph}$
(d) For large potential difference $(V > > \phi / e), \lambda _e$ is approximately halved if $V$ is made four times.
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Answer:
Correct Answer: 181.(d)
Solution:
Formula:
- $K _{\text {max }}=\frac{h c}{\lambda _{ph}}-\phi$
Kinetic energy of electron reaching the anode will be
$ K=\frac{h c}{\lambda _{ph}}-\phi+e V $
Now, $\lambda _e=\frac{h}{\sqrt{2 m K}}=\frac{h}{\sqrt{2 m (\frac{h c}{\lambda _{ph}}-\phi+e V)}}$
$ \begin{aligned} & \text { If } eV \gg \phi \text { then, } \lambda _e=\frac{h}{\sqrt{2 m (\frac{h c}{\lambda _{ph}}+e V)}} \\ & \text { If } V _f=4 V _i \text { then, }\left(\lambda _e\right) _f \simeq \frac{\left(\lambda _e\right) _i}{2} \end{aligned} $
BETA
