A proton, a neutron, an electron and an α -particle have same energy. If λ p , λ n , λ e…

The de-Broglie wavelength $\mathrm{\lambda } = \frac{\mathrm{h}}{\sqrt{2\mathrm{m}(\mathrm{K}.\mathrm{E}.)}}$

where, $\mathrm{h}$ and $\mathrm{K}.\mathrm{E}.$ are constant

$\mathrm{\lambda }\propto \frac{1}{\sqrt{\mathrm{m}}}$

$\mathrm{m}=$ Mass of the particle

Thus, with increasing mass of particle de-Broglie wavelength decreases.

Order of mass $m_e<m_p<m_n<m_{\alpha }$

Order of de-Broglie wavelength is ${\lambda }_a<{\lambda }_n<{\lambda }_p<{\lambda }_e$