A proton and an α-particle are accelerated from rest by 2 V and 4 V potentials, respectively. The ratio of their de-Broglie wavelength is:4 : 12 : 18 : 116 : 1

The de-Broglie wavelength is given by the formula:

λ = h / p

where h is Planck's constant and p is the momentum of the particle.

The momentum of a particle is given by the formula:

p = sqrt(2mE)

where m is the mass of the particle and E is the energy of the particle.

The energy of a particle is given by the formula:

E = qV

where q is the charge of the particle and V is the potential difference.

For a proton, q = e (the elementary charge) and m = m_p (the mass of a proton). For an α-particle, q = 2e and m = 4m_p.

So, the de-Broglie wavelength of a proton is:

λ_p = h / sqrt(2m_p * e * 2V)

And the de-Broglie wavelength of an α-particle is:

λ_α = h / sqrt(2 * 4m_p * 2e * 4V)

Taking the ratio of these two wavelengths gives:

λ_p / λ_α = sqrt((2 * 4m_p * 2e * 4V) / (2m_p * e * 2V)) = sqrt(16/2) = 2

So, the ratio of the de-Broglie wavelengths of a proton and an α-particle is 2:1.