The wavelength λ_e of an electron and λ_P of a photon of same...

For a photon of energy $E$, the wavelength is given by:

$\lambda_P = \dfrac{hc}{E}$

For an electron of kinetic energy $E$, the de Broglie wavelength is given by:

$\lambda_e = \dfrac{h}{\sqrt{2mE}}$

Squaring both sides, we get:

$\lambda_e^2 = \dfrac{h^2}{2mE}$

$E = \dfrac{h^2}{2m\lambda_e^2}$

Substituting the value of $E$ in the expression for $\lambda_P$:

$\lambda_P = \dfrac{hc}{\left(\dfrac{h^2}{2m\lambda_e^2}\right)}$

$\lambda_P = \left(\dfrac{2mc}{h}\right) \lambda_e^2$

Since $m$, $c$, and $h$ are constants, we obtain:

$\lambda_P \propto \lambda_e^2$

Answer: $\lambda_P\ \alpha\ \lambda_e^2$