A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_{0}$ . 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: (in the usual notation)

\frac{3 m (E - W_{0})}{e B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{2 m (E - W_{0}) e B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{\sqrt{2 e (E - W_{0})}}{m B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{\sqrt{2 m (E - W_{0})}}{e B}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{\sqrt{2 m (E - W_{0})}}{e B}$
Kinetic energy of photoelectron $K E = E - W_{0}$

Radius of circular path when electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field is
$r = \frac{m v}{e B} = \frac{p}{e B} = \frac{\sqrt{2 m K E}}{e B} = \frac{\sqrt{2 m (E - W_{0})}}{e B}$

Suggest Corrections

Similar questions

W_{0} .

W_{0}

λ_{0}

\frac{W_{0}}{2}

W_{0}

λ_{0}

\frac{W_{0}}{2}

y = A sin (98 t + w_{0}) + B cos (98 t + w_{0})

w_{0}

y_{2} + C y = 0

R

Join BYJU'S Learning Program