A proton and an alpha particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is

CONCEPT:

• de Broglie wavelength of electrons: Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
  • The wavelength of material waves is also known as the de Broglie wavelength.
• de Broglie wavelength of electrons can be calculated from Planks constant h divided by the momentum of the particle.

λ = h/p

where λ is de Broglie wavelength, h is Plank's const, and p is the momentum.

In terms of Energy, de Broglie wavelength of electrons:

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

• Electric Potential energy: If a Charge q is placed in the electric potential V, the electric potential energy

U = E = qV

where U or E is the electric potential energy, q is the charge, and V is the electric potential.

CALCULATION:

m_p and q_p is the mass and charge of the proton.

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

4m_p =  mα  and 2q_p = qα

So, the correct answer is option 1.