Question

A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is ${{W}_{o}}$ . If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$ , then the radius $r$ is given by A. $\left( \text{in the usual notation} \right)$
A. $\sqrt{\dfrac{2m\left( E-{{W}_{o}} \right)}{eB}}$
B. $\sqrt{2m\left( E-{{W}_{o}} \right)eB}$
C. $\dfrac{\sqrt{2m\left( E-{{W}_{o}} \right)}}{mB}$
D. $\dfrac{\sqrt{2m\left( E-{{W}_{o}} \right)}}{eB}$

Answer 1

When a photoelectron is ejected out of a metal surface, it bears maximum Kinetic energy which can be given by $K{{E}_{\max }}=E-{{W}_{o}}$where $E$is the photon energy and${{W}_{o}}$is the binding energy or Work function of the electron to the particular material .
And when an electron enters a region of magnetic induction$B$, it experiences a a force exerted by the magnetic field describing a circular path of radius $r$and this force is generally centrifugal force which can be given by$F=\dfrac{m{{v}^{2}}}{r}$.

Complete step by step answer:
The two forces being worked on,here, are the magnetic force working on an electron $F=evB$ and centrifugal force $F=\dfrac{m{{v}^{2}}}{r}$ which must be equated to find the radius of the circular path $r$ and substitute the velocity found from the equation $E={{W}_{o}}+K{{E}_{\max }}$ .
We know that the magnetic force $F=e\left( \overrightarrow{V}\times \overrightarrow{B} \right)$
$\Rightarrow F=evB$
Where $F=\text{force}$
$e=\text{charge of electron}$
$v=\text{velocity}$
$B=\text{Strength of the magnetic field}$
This is equal to the centrifugal force, that is $F=\dfrac{m{{v}^{2}}}{r}$
By equating these two forces we finally have the expression for the radius of the circular path as
$r=\dfrac{mv}{eB}$
We also know that $E={{W}_{o}}+K{{E}_{\max }}$
$\Rightarrow K{{E}_{\max }}=E-{{W}_{o}}$
We know the relation between the momentum and the kinetic energy as below
$mv=\sqrt{2mK{{E}_{\max }}}$
Put the kinetic energy expression we have above
$mv=\sqrt{2m\left( E-{{W}_{o}} \right)}$
In the above equation of $r$, substitute the momentum term
$r=\dfrac{\sqrt{2m\left( E-{{W}_{o}} \right)}}{eB}$
This is the required final expression for the radius of the circular path.
Therefore the expression for the radius of the circular path is $r=\dfrac{\sqrt{2m\left( E-{{W}_{o}} \right)}}{eB}$
Hence option $\left( D \right)$ is correct.

Note:
One must carefully observe how in such a situation magnetic force is balanced by the centrifugal force of the photon moving in the circular path, following the universal conservation law of energy, and Hence to find the expression for the radius of the circular path, both the forces are equated.