QJEEAdv16Pha231: JEE Advanced 2016 Physics - Class 12 JEEAdv16Pha231 Solution: Light of wavelength ph falls on a cathode plate inside a vacuum …

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,

${\lambda _e} = {h \over {\sqrt {2m\,{K_{\max ,\,a}}} }} = {h \over {\sqrt {2m\left( {{{hc} \over {{\lambda _{ph}}}} - \phi + eV} \right)} }}$ ...... (i)

D : For V > > \phi e

${\lambda _e} \approx {h \over {\sqrt {2meV} }} \propto {1 \over {\sqrt V }}$

$\therefore$ 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)

${{d{\lambda _e}} \over {dt}} = {{ - h} \over 2}{\left[ {2m\left( {{{hc} \over {{\lambda _{ph}}}} - \phi + eV} \right)} \right]^{ - 3/2}} \times 2m\left( {{{ - hc} \over {\lambda _{ph}^2}}} \right){{d{\lambda _{ph}}} \over {dt}}$

$\Rightarrow {{d{\lambda _e}} \over {dt}} \ne {{d{\lambda _{ph}}} \over {dt}}$

I : $\lambda$ _e does not depend on d.