Question

In a photo-emissive cell with exciting wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $\dfrac{3\lambda }{4}$ the speed of the fastest emitted electron will be:
A. $v{{\left( 3/4 \right)}^{1/2}}$
B. $v{{\left( 4/3 \right)}^{1/2}}$
C. Less than $v{{\left( 4/3 \right)}^{1/2}}$
D. Greater than $v{{\left( 4/3 \right)}^{1/2}}$

Answer 1

Here first we will know what does photo-emissive cell mean and then we will derive the maximum velocity of the electron and based on that we will find the speed of the fastest emitted electron meeting all conditions given in the problem.

Formula used:
Photoelectric effect equation is given by:
$E={{W}_{0}}+\dfrac{1}{2}m{{v}^{2}}_{\max }$
Where ${{W}_{0}}$ is the energy required to eject an electron (work function), $m$ is the mass of the electron and ${{v}_{\max }}$ is the maximum velocity of the electron.

Complete step by step answer:
A phototube or photoelectric cell is a type of gas-filled or a vacuum tube which is sensitive to light. Such a tube is more accurately called a photo-emissive cell to distinguish it from photovoltaic or photoconductive cells. In other words when light strikes materials, it can eject electrons from them. This is called the photoelectric effect which means that light (photo) produces electricity. As we know that photoelectric effect equation is given by:
$E={{W}_{0}}+\dfrac{1}{2}m{{v}^{2}}_{\max }$

By rearranging
${{v}_{\max }}=\sqrt{\dfrac{2E}{m}-\dfrac{2{{W}_{0}}}{m}}$
Where $E=\dfrac{hc}{\lambda }$ which is the photon energy meaning the amount of energy a photon has. When the wavelength of incident light changes from $\lambda$ to $\dfrac{3\lambda }{4}$(when it decreases). Let energy of incident light charges from $E$ and speed of fastest electron changes from $v$ to $v'$ then
$v'=\sqrt{\dfrac{2E'}{m}-\dfrac{2{{W}_{0}}}{m}}$
As $E\propto \dfrac{1}{\lambda }\Rightarrow E'\dfrac{4}{3}E$
$v'=\sqrt{\dfrac{2\left( \dfrac{4}{3} \right)E}{m}-\dfrac{2{{W}_{0}}}{m}}$
$\Rightarrow v'={{\left( \dfrac{4}{3} \right)}^{1/2}}\sqrt{\dfrac{2E}{m}-\dfrac{2{{W}_{0}}}{m{{\left( 4/3 \right)}^{1/2}}}}$
$\therefore v'>{{\left( 4/3 \right)}^{1/2}}v$

Therefore, the speed of the fastest emitted electron will be ${{\left( 4/3 \right)}^{1/2}}$ times the original speed.

Note: One must take note that at threshold frequency, electrons are just ejected and so they do not have any kinetic energy and below threshold frequency there is no emission of electrons. Therefore, the energy of an electron with this frequency is equal to the work function of the metal.