The de Broglie wavelengths for an electron and a photon are $$\lambda_e$$ and $$\lambda_p$$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two?

We need to find the relation between de Broglie wavelengths of an electron ($$\lambda_e$$) and a photon ($$\lambda_p$$) when both have the same kinetic energy $$K$$. For an electron with kinetic energy $$K$$, $$K = \frac{p_e^2}{2m_e}$$, so $$p_e = \sqrt{2m_eK}$$. Then $$\lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_eK}}$$, which gives $$K = \frac{h^2}{2m_e\lambda_e^2}$$.

For a photon with energy $$K$$, $$K = \frac{hc}{\lambda_p}$$ and hence $$\lambda_p = \frac{hc}{K}$$. Substituting the expression for $$K$$ from the electron case into this yields

$$\lambda_p = \frac{hc}{\frac{h^2}{2m_e\lambda_e^2}} = \frac{hc \times 2m_e\lambda_e^2}{h^2} = \frac{2m_ec\lambda_e^2}{h}$$.

Since $$\frac{2m_ec}{h}$$ is a constant, $$\lambda_p \propto \lambda_e^2$$. The correct answer is Option A.