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

Answer: (D)

$K_p$ $<$ $K_e$

Answer 2

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:
$K_p<K_e.$