The de Broglie wavelengths of an electron, a proton, a neutron, and an alpha-particle are represented as λₑ, λₚ, λₙ and λₐ respectively. If all these particles have same kinetic energies, then which one of the following relations is correct?

Correct option is D

Correct answer is D
Calculation:
Kinetic energy (KE) = p2/2m ⇒ p = √(2m × KE).
For the same KE, momentum p increases with √m.
From λ = h/p, λ decreases as p increases.
Mass order: me (electron) < mp (proton) ≃ mn (neutron) < mα (alpha particle).
Therefore, wavelength order: λe > λp ≃ λn > λα.