A proton, a neutron, an electron and an α -particle have same energy. Then, their de-Broglie wavelengths compare as:

$λ_α<λ_p=λ_n<λ_e$

We know that the relation between λ and K is given by $λ=\frac{h}{\sqrt{2mK}}$

Here, for the given value of energy K, $\frac{h}{\sqrt{2K}}$ is a constant.

Thus, $λ∝\frac{1}{\sqrt{m}}$

$∴λ_p:λ_n:λ_e:λ_α$

$⇒=\frac{1}{\sqrt{m_p}}:\frac{1}{\sqrt{m_n}}:\frac{1}{\sqrt{m_e}}:\frac{1}{\sqrt{m_α}}$

Since, $m_p=m_n$, hence $λ_p=λ_n$

As, $m_α<m_p$, therefore $λ_α>λ_n$

Hence, $λ_α<λ_p=λ_n<λ_e$