Final answer:
The ratio of the de Broglie wavelengths of a proton and alpha-particle accelerated through the same potential difference is 2 because the momentum of the alpha particle is twice that of the proton due to its greater mass and charge. Thus, the de Broglie wavelength of the proton is twice that of the alpha particle.
Step-by-step explanation:
The subject of this question is de Broglie wavelength of particles in Physics, particularly within the scope of quantum mechanics. The grade level would be High School or introductory college courses, where topics like the de Broglie hypothesis and the behavior of subatomic particles are discussed.
To find the ratio of the de Broglie wavelengths of a proton and an alpha-particle, we will use the de Broglie wavelength formula λ = h/p, where λ is the de Broglie wavelength, h is the Planck constant, and p is the momentum of the particle. Since both particles are accelerated through the same potential difference, they will have gained the same amount of kinetic energy (K.E = qV, where q is the charge and V is the potential difference). The momentum p of a particle can be related to its kinetic energy by p = √(2mK.E), where m is the mass of the particle.
An alpha particle has four times the mass of a proton and twice its charge. Therefore, since kinetic energy K.E acquired from the potential difference is equal, for the alpha particle, pα = √(2(4m_p)(q_pV)) and for the proton, p_p = √(2m_p(q_pV)). Now, let's find the ratio of their momenta p_p/pα, which simplifies to 1/2. Since the de Broglie wavelength is inversely proportional to momentum, the ratio of the wavelengths λ_p / λα is the inverse of the momenta ratio, hence 2. So the correct answer is A. 2.