A photon of energy ' $E$ ' ejects photoelectrons from a metal surface whose work function is $W_0$. If this…
A photon of energy ' $E$ ' ejects photoelectrons from a metal surface whose work function is $W_0$. If this electron enters into a uniform magnetic field with induction ' $B$ ' in a direction perpendicular to the field and describes a circular path of radius ' $r$ ', then the radius is given by
$\sqrt{\frac{2 e\left(E-W_0\right)}{m B}}$
$\frac{\sqrt{2\left(E-W_0\right) m}}{e B}$
$\sqrt{\frac{2 m\left(E-W_0\right)}{m B}}$
$\sqrt{2 m\left(E-W_0\right) e B}$
Solution
Concept: Photoelectric effect and the motion of charge in a direction perpendicular to the magnetic field`
The maximum kinetic energy $K=\frac{m v^2}{2}$ of a photo-electron is given by,'
$K=E-W_0$
where, $E$ is the energy of incident photon and $W_0$ is the work function of the metal.
Therefore, the velocity of the photoelectron is: $v=\sqrt{\frac{2\left(E-W_0\right)}{m}}$
The photoelectron undergoes uniform circular motion as its velocity is perpendicular to the magnetic field. On balancing magnetic force on moving charge with centrifugal force:
$q v B=\frac{m v^2}{r}$
Therefore, $r=\frac{m v}{B q}=\frac{\sqrt{2\left(E-W_0\right) m}}{e B}$