Ratio of Change in De Broglie Wavelength to Incident Wavelength

To find the relation for the ratio $\left| \frac{\Delta \lambda_d}{\Delta \lambda} \right|$, we apply Einstein's photoelectric equation and the de Broglie wavelength formula.

According to Einstein's photoelectric equation, the maximum kinetic energy $K_{\max}$ of an emitted photoelectron is given by: $K_{\max} = \frac{hc}{\lambda} - \phi_0$

The de Broglie wavelength $\lambda_d$ of the fastest emitted photoelectron is related to its maximum kinetic energy $K_{\max}$ by: $\lambda_d = \frac{h}{p_{\max}} = \frac{h}{\sqrt{2m K_{\max}}}$

Squaring both sides and expressing $\lambda_d$ in terms of $\lambda$: $\lambda_d^2 = \frac{h^2}{2m \left(\frac{hc}{\lambda} - \phi_0\right)}$

Taking the reciprocal of both sides: $\lambda_d^{-2} = \frac{2m}{h^2} \left(\frac{hc}{\lambda} - \phi_0\right)$

To find the relationship between small changes $\Delta \lambda_d$ and $\Delta \lambda$, we differentiate both sides with respect to $\lambda$: $-2 \lambda_d^{-3} \frac{d\lambda_d}{d\lambda} = \frac{2m}{h^2} \left(-\frac{hc}{\lambda^2}\right)$

Simplifying the equation: $\frac{1}{\lambda_d^3} \frac{d\lambda_d}{d\lambda} = \frac{mc}{h \lambda^2}$

Rearranging for the ratio of the differentials: $\frac{d\lambda_d}{d\lambda} = \left(\frac{mc}{h}\right) \frac{\lambda_d^3}{\lambda^2}$

For small variations $\Delta \lambda_d$ and $\Delta \lambda$: $\left| \frac{\Delta \lambda_d}{\Delta \lambda} \right| = \left(\frac{mc}{h}\right) \frac{\lambda_d^3}{\lambda^2}$

Since $m$, $c$, and $h$ are physical constants, we conclude that: $\frac{\Delta \lambda_d}{\Delta \lambda} \propto \frac{\lambda_d^3}{\lambda^2}$

Correct Answer: D ($\lambda_d^3/\lambda^2$)