Question

A proton and an electron have the same de Broglie wavelength. If \(K_p\) and \(K_e\) be the kinetic energies of proton and electron respectively. Then choose the correct relation:

• A. \(K_p\) \(>\) \(K_e\)
• B. \(K_p = K_e\)
• C. \(K_p = K_e^2\)
• D. \(K_p\) \(<\) \(K_e\)

Answer 1

The Correct Option is D

The de Broglie wavelength for a particle is given by:
\[\lambda = \frac{h}{p},\]
where \(h\) is Planck's constant and \(p\) is the momentum.
For the proton and electron:
\[\lambda_{\text{proton}} = \lambda_{\text{electron}} \implies p_{\text{proton}} = p_{\text{electron}}.\]
The kinetic energy is related to momentum as:
\[K = \frac{p^2}{2m}.\]
Since \(p_{\text{proton}} = p_{\text{electron}}\):
\[K_{\text{proton}} = \frac{p^2}{2m_p}, \quad K_{\text{electron}} = \frac{p^2}{2m_e}.\]
Given \(m_p>m_e\), it follows that:
\[K_{\text{proton}}<K_{\text{electron}}.\]
Thus:
\[K_p<K_e.\]