Ratio of De-Broglie wavelengths of a proton and an alpha particle of the same energy is?

The de Broglie wavelength of electrons can be determined using Plank's constant, h, divided by the particle's momentum.

λ = h/p

where h = Plank's constant, p = momentum, and  λ = de Broglie wavelength.

De Broglie wavelength of a charge particle, in terms of energy:

λ = h/√2mE

where λ is de Broglie wavelength, h is Plank's const, m is the mass of the particle, and E is the energy of the particle.

mp and qp are the mass and charge of the proton, respectively

mα and qα is the mass and charge of the alpha particle, respectively.

Energy (E) is the same for both:

4mp = mα and 2qp = qα

Since λ = h/√2mE

λp/λα = √mα/mp = √(4mp/mp) = 2

Thus ratio = 2 :1

Summary:

Ratio of De-Broglie wavelengths of a proton and an alpha particle of the same energy is? (A) 1 : 4 (B) 4 : 1 (C) 1 : 2 (D) 2 : 1

The ratio of the De-Broglie wavelengths of an alpha particle and a proton with the same energy is 2:1.