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(D) According to Einstein's photoelectric equation, the kinetic energy $K$ of the emitted photoelectron is given by $K = \frac{hc}{\lambda_i} - \frac{

Vedclass · Accepted Answer

$[h/[2mc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})]]^{1/2}$

Vedclass · Answer

(D) According to Einstein's photoelectric equation, the kinetic energy $K$ of the emitted photoelectron is given by $K = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0}$.Substituting $K = \frac{p^2}{2m}$, where $p$ is the momentum of the electron, we get $\frac{p^2}{2m} = hc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})$.From the de-Broglie relation, the wavelength $\lambda_e$ is given by $\lambda_e = \frac{h}{p}$, which implies $p = \frac{h}{\lambda_e}$.Substituting $p$ in the kinetic energy equation: $\frac{h^2}{2m\lambda_e^2} = hc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})$.Rearranging for $\lambda_e^2$: $\lambda_e^2 = \frac{h^2}{2mhc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})} = \frac{h}{2mc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})}$.Thus, $\lambda_e = [\frac{h}{2mc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})}]^{1/2}$.

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