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

\(\lambda \)_{\(\alpha \)} < \(\lambda \)p = \(\lambda \)n < \(\lambda \)e

Since, λ ∝ \(\frac{1}{\sqrt{ m}}\)

and the order of mass of these particles are m_e < m_p = m_n < m\(\alpha \)

so, the correct order of their wavelengths will be

\(\lambda \)_{\(\alpha \)} < \(\lambda \)_p = \(\lambda \)_n < \(\lambda \)_e