A photon of energy E ejects a photoelectron f | vedclass

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Question

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the ejected photoelectron is given by:$K_{max} = E - W_0 =

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

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

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