A photon of energy E ejects a photoelectron f | vedclass

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Question

(A) 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

Vedclass · Accepted Answer

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

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(A) 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 in a uniform magnetic field $B$ perpendicular to its velocity,it experiences a magnetic Lorentz force which provides the necessary centripetal force for circular motion:$evB = \frac{mv^2}{r}$Solving for the radius $r$,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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