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:$K = E - W_0 = \frac{1}{2}mv^2$From

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

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(A) According to Einstein's photoelectric equation,the kinetic energy $K$ of the ejected photoelectron is given by:$K = E - W_0 = \frac{1}{2}mv^2$From this,the velocity $v$ of the electron is:$v = \sqrt{\frac{2(E - W_0)}{m}}$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:$evB = \frac{mv^2}{r}$Rearranging for the radius $r$:$r = \frac{mv}{eB}$Substituting the expression for $v$:$r = \frac{m}{eB} \sqrt{\frac{2(E - W_0)}{m}} = \frac{\sqrt{2m(E - W_0)}}{eB}$

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