Question 55: Dual Nature of Matter and Radiation

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 statement(s) is (are) true?

Accepted Answer

In photo-electric effect, the maximum kinetic energy of the photo-electron ejected at the cathode, is given by
$${K_{\max ,\,c}} = {{hc} \over {{\lambda _{ph}}}} - \phi $$,
where $$\lambda$$
ph
is the wavelength of the incident light and $$\phi$$ is the work function of the material. The ejected photoelectrons are accelerated from the cathode to the anode by a potential V. Thus, the maximum kinetic energy of the photo-electron at the anode is
$${K_{\max ,\,a}} = {K_{\max ,\,c}} + eV = {{hc} \over {{\lambda _{ph}}}} - \phi  + eV$$.
Now, minimum de-Broglie wavelength of electrons passing through anode,
$${\lambda _e} = {h \over {\sqrt {2m\,{K_{\max ,\,a}}} }} = {h \over {\sqrt {2m\left( {{{hc} \over {{\lambda _{ph}}}} - \phi  + eV} ight)} }}$$ ...... (i)
D : For $$V >  > \phi e$$
$${\lambda _e} \approx {h \over {\sqrt {2meV} }} \propto {1 \over {\sqrt V }}$$
$$	herefore$$ If V is made four times, $$\lambda$$
e
becomes $${1 \over {\sqrt 4 }} = {1 \over 2}$$, i.e. halved.
C : When $$\phi$$ and $$\lambda$$
ph
increases, $$\lambda$$
e
also increases.
A : From (i)
$${{d{\lambda _e}} \over {dt}} = {{ - h} \over 2}{\left[ {2m\left( {{{hc} \over {{\lambda _{ph}}}} - \phi  + eV} ight)} ight]^{ - 3/2}} 	imes 2m\left( {{{ - hc} \over {\lambda _{ph}^2}}} ight){{d{\lambda _{ph}}} \over {dt}}$$
$$ \Rightarrow {{d{\lambda _e}} \over {dt}} 
e {{d{\lambda _{ph}}} \over {dt}}$$
I : $$\lambda$$
e
does not depend on d.

Answer

$$\lambda$$e increases at the same rate as $$\lambda$$ph for $$\lambda$$ph < hc/$$\phi$$

Answer

$$\lambda$$e is approximately halved, if d is doubled

Answer

$$\lambda$$e decreases with increase in $$\phi$$ and $$\lambda$$ph

Answer

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

Dual Nature of Matter and Radiation

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