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 statement(s) is(are) true?

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

Answer 1

The correct option is D For large potential difference $(V>>\varphi /e){\lambda }_e$ is approximately halved if V is made four time

Electrons entering the cathode having minimum de broglie wavelength ${\lambda }_e$.
So kinetic energy of those elctrons will be maximum

$K.E_{max}=\frac{hc}{{\lambda }_{ph}}-\varphi$.

Kinetic energy of electrons entering from anode is $K.E_{anode}=\frac{hc}{{\lambda }_{ph}}-\varphi +eV=\frac{p^2}{2m_e}=\frac{h^2}{2m_e{\lambda }_e^2}$
$\therefore$ $K.E_{anode}=\frac{hc}{{\lambda }_{ph}}-\varphi +eV=\frac{h^2}{2m_e{\lambda }_e^2}$

diffrentiating both the sides,
$\frac{-hc}{{\lambda }_{ph}^2}\times d{\lambda }_{ph}$ $=$ $\frac{-2h^2}{2m_e{\lambda }_e^3}\times d{\lambda }_e$
$d{\lambda }_{ph}=(\frac{h}{m_{e}c}\times \frac{{\lambda }_{ph}^2}{{\lambda }_e^3})$.$d{\lambda }_e$
Here we can say that option A is wrong,
and ${\lambda }_e$ does not depends on the distance 'd' which is distance between anode and cathode as long as potential diffrence is same, hence option B is wrong.

Here if we increase $\varphi$ and ${\lambda }_{ph}$ the kinetic energy will reduce as
$K.E_{anode}=\frac{hc}{{\lambda }_{ph}}-\varphi$
and $\therefore$ ${\lambda }_e$ will increase, so the option C is also incorrect.

$V>>\frac{\varphi }{e}$

Energy of electron,

$eV\approx \frac{h^2}{2m_e{\lambda }_e^2}$

This supports option D.