A photon of energy E ejects a photoelectron from a metal surface whose work function is `phi_(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)

Options

A. `sqrt((2m(E-phi_(0)))/(eB))`
B. `sqrt((2m(E-phi_(0)))eB)`
C. `sqrt((2m(E-phi_(0)))/(mB))`
D. `sqrt(2m(E-phi_(0)))/(eB)`

Answer

D

Explanation

(d) : As the electron describes a circular path of radius r in the magnetic field, therefore
`(mv^(2))/(r)=evB`
`r=(mv)/(eB)=(p)/(eB)=sqrt(2mK)/(eB)" "("As K"=(p^(2))/(2m))`
From Einstein's photoelectric equation
`K=E-phi_(0):.(sqrt(2m(E-phi_(0))))/(eb)`