A photon of energy E ejects a photoelectron f | vedclass

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(D) According to Einstein's photoelectric equation,the kinetic energy $K$ of the ejected photoelectron is given by $K = E - W_0$.Since $K = \frac{1}{2

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$\frac{\sqrt{2m(E - W_0)}}{eB}$

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(D) According to Einstein's photoelectric equation,the kinetic energy $K$ of the ejected photoelectron is given by $K = E - W_0$.Since $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2(E - W_0)}{m}}$.When a charged particle of mass $m$ and charge $e$ moves with velocity $v$ perpendicular to a uniform magnetic field $B$,it experiences a magnetic Lorentz force that provides the necessary centripetal force for circular motion.Thus,$evB = \frac{mv^2}{r}$.Solving for the radius $r$,we get $r = \frac{mv}{eB}$.Substituting the expression for $v$ into the equation for $r$:$r = \frac{m}{eB} \sqrt{\frac{2(E - W_0)}{m}} = \frac{\sqrt{m^2 \cdot \frac{2(E - W_0)}{m}}}{eB} = \frac{\sqrt{2m(E - W_0)}}{eB}$.

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