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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B), (A), (C), (D)
$\text{Given: A proton, an electron, a neutron, and an alpha particle have the same energy.}$
$\text{De Broglie wavelength: } \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m E}}$
$\text{Since } E \text{ is same, } \lambda \propto \frac{1}{\sqrt{m}}$
$m_\text{electron} \ll m_\text{proton} \approx m_\text{neutron} \ll m_\text{alpha}$
$\text{Thus, } \lambda_\text{electron} > \lambda_\text{proton} \approx \lambda_\text{neutron} > \lambda_\text{alpha}$