A proton, an electron, a neutron and an alpha particle have the same energy. Their de Broglie's wavelengths are given below

(A) $λ_{\text{proton}}$
(B) $λ_{\text{electron}}$
(C) $λ_{\text{neutron}}$
(D) $λ_{\text{alpha}}$

Arrange these wavelengths in decreasing order.

Choose the correct answer from the options given below:

(B), (A), (C), (D)

The correct answer is Option (2) → (B), (A), (C), (D)

$\text{Given: A proton, an electron, a neutron, and an alpha particle have the same energy.}$

$\text{De Broglie wavelength: } \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m E}}$

$\text{Since } E \text{ is same, } \lambda \propto \frac{1}{\sqrt{m}}$

$m_\text{electron} \ll m_\text{proton} \approx m_\text{neutron} \ll m_\text{alpha}$

$\text{Thus, } \lambda_\text{electron} > \lambda_\text{proton} \approx \lambda_\text{neutron} > \lambda_\text{alpha}$