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Step-by-Step Solution
Step 1: Equating de Broglie Wavelengths
Given that the proton and electron have the same de Broglie wavelength, we use the de Broglie relation:
$ \lambda = \frac{h}{p}. $
Hence, for the proton ($\lambda_p$) and for the electron ($\lambda_e$):
$ \lambda_p = \frac{h}{P_p}, \quad \lambda_e = \frac{h}{P_e}. $
Since $ \lambda_p = \lambda_e,$ it follows that:
$ P_p = P_e. $
Thus, the proton and electron must have the same momentum if their de Broglie wavelengths are the same.
Step 2: Expressing Kinetic Energies
The kinetic energy $K$ of a particle of momentum $p$ and mass $m$ is given by:
$ K = \frac{p^2}{2m}. $
So, for the proton:
$ K_p = \frac{P_p^2}{2 m_p}, $
and for the electron:
$ K_e = \frac{P_e^2}{2 m_e}. $
Since $P_p = P_e = P,$ we can write:
$ K_p = \frac{P^2}{2 m_p}, \quad K_e = \frac{P^2}{2 m_e}. $
Step 3: Comparing Kinetic Energies
Because the mass of the proton $m_p$ is greater than the mass of the electron $m_e$ (i.e. $m_p > m_e$), for the same value of $P^2,$ the denominator $2m_p$ is larger than $2m_e.$ Therefore:
$ K_p = \frac{P^2}{2 m_p} < \frac{P^2}{2 m_e} = K_e. $
Hence, $K_p < K_e.$
Step 4: Final Conclusion
The momentum of the proton and electron are equal, but the kinetic energy of the proton is less than that of the electron. Therefore, the correct choice is:
$ K_p < K_e \quad \text{and} \quad P_p = P_e. $
