A photoelectric material having work-function | vedclass

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the photoelectron is given by:$K_{max} = \frac{hc}{\lambda}

Vedclass · Accepted Answer

$\lambda_d^3 / \lambda^2$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the photoelectron is given by:$K_{max} = \frac{hc}{\lambda} - \phi_0$Since $K_{max} = \frac{p^2}{2m}$ and the de Broglie wavelength is $\lambda_d = \frac{h}{p}$,we have $p = \frac{h}{\lambda_d}$.Substituting this into the kinetic energy equation:$\frac{h^2}{2m \lambda_d^2} = \frac{hc}{\lambda} - \phi_0$Differentiating both sides with respect to $\lambda$ and $\lambda_d$:$\frac{d}{d\lambda_d} \left( \frac{h^2}{2m \lambda_d^2} \right) d\lambda_d = \frac{d}{d\lambda} \left( \frac{hc}{\lambda} - \phi_0 \right) d\lambda$$\frac{h^2}{2m} (-2 \lambda_d^{-3}) d\lambda_d = -hc \lambda^{-2} d\lambda$$\frac{h^2}{m \lambda_d^3} d\lambda_d = \frac{hc}{\lambda^2} d\lambda$Rearranging for the ratio $\frac{d\lambda_d}{d\lambda}$:$\frac{d\lambda_d}{d\lambda} = \frac{hc}{\lambda^2} \cdot \frac{m \lambda_d^3}{h^2} = \left( \frac{mc}{h} \right) \frac{\lambda_d^3}{\lambda^2}$Thus,$\frac{\Delta \lambda_d}{\Delta \lambda} \propto \frac{\lambda_d^3}{\lambda^2}$.

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