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Question

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)$ of the emitted photoelectrons is given by: $K.E. = E - W$,where

Vedclass · Accepted Answer

$\lambda_c = \sqrt{\frac{h}{2mc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}}$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)$ of the emitted photoelectrons is given by: $K.E. = E - W$,where $E = \frac{hc}{\lambda_i}$ is the energy of the incident photon and $W = \frac{hc}{\lambda_0}$ is the work function of the substance.Substituting these,we get: $K.E. = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} = hc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)$.The de-Broglie wavelength $\lambda_c$ of an electron with kinetic energy $K.E.$ is given by $\lambda_c = \frac{h}{\sqrt{2m(K.E.)}}$.Squaring both sides,we get $\lambda_c^2 = \frac{h^2}{2m(K.E.)}$,which implies $K.E. = \frac{h^2}{2m\lambda_c^2}$.Equating the two expressions for $K.E.$: $\frac{h^2}{2m\lambda_c^2} = hc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)$.Solving for $\lambda_c$: $\lambda_c^2 = \frac{h^2}{2mhc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)} = \frac{h}{2mc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}$.Therefore,$\lambda_c = \sqrt{\frac{h}{2mc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}}$.

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