A proton, a neutron, an electron and an \(\alpha \)-particle have same energy. Then their de-Broglie wavelengths compare as :
Answer & explanation
Correct answer: option 2
Since, λ ∝ \(\frac{1}{\sqrt{ m}}\)
and the order of mass of these particles are me < mp = mn < m\(\alpha \)
so, the correct order of their wavelengths will be
\(\lambda \)\(\alpha \) < \(\lambda \)p = \(\lambda \)n < \(\lambda \)e