If velocity of a particle is three times that of an electron and the ratio of the de Broglie wavelength of the particle to that of the electron is 1.814 × 10⁻⁴, what is the particle?

The correct option is A Neutron
de Broglie wavelength, λ = h/p ⇒ λ = h/mv ⇒ λ ∝ 1/v
⇒ λp/λe = meve/mpvp
⇒ mp = me(ve/vp) × (λe/λp)
Here, me = 9.1 × 10⁻³¹ kg, vp = 3ve and λp/λe = 1.814 × 10⁻⁴
⇒ mp = (9.1 × 10⁻³¹)/(1.814 × 10⁻⁴) × 3 = 1.672 × 10⁻²⁷ kg
Thus, the particle is a neutron.