1. Use de-Broglie relation

For any particle, $\lambda = \frac{h}{p}$ where $\lambda$ is de-Broglie wavelength and $p$ is momentum.

Since the proton and electron have the same de-Broglie wavelength, we must have $\lambda_p = \lambda_e$ So, $\frac{h}{P_p} = \frac{h}{P_e} \implies P_p = P_e$

Thus, the momenta are equal.

2. Relate kinetic energy to momentum

For a non-relativistic particle, $K = \frac{p^2}{2m}$

Since both particles have the same momentum, $K \propto \frac{1}{m}$

Now, $m_p \gg m_e$ So for the same $p$, $K_p = \frac{p^2}{2m_p}, \qquad K_e = \frac{p^2}{2m_e}$

Because $m_p > m_e$, $K_p < K_e$

3. Check options

• A: $K_p < K_e$ and $P_p = P_e$ ✅
• B: $K_p = K_e$ and $P_p = P_e$ ❌
• C: $K_p < K_e$ and $P_p < P_e$ ❌
• D: $K_p > K_e$ and $P_p = P_e$ ❌

4. Final answer

The correct option is: $\boxed{A}$