An electron of mass mₑ and a proton of mass mₚ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

Solution

Physics sir · 3 steps

Same voltage means same kinetic energy eV for both, so the wavelength depends only on mass.

1. Same energy gained. A charge e accelerated through a potential difference V gains kinetic energy eV. The electron and the proton carry charges of the same size, so both end up with the same kinetic energy K = eV.
2. Wavelength from energy. Momentum is p = √2mK, and the de Broglie wavelength is λ = h/p. Since h, e and V are the same for both particles, λ ∝ 1/√m.

   λ = h/√2meV ⇒ λₑ/λₚ = √(mₚ/mₑ)

3. So. The lighter electron has the longer wavelength, by a factor of √(mₚ/mₑ). The answer is (d).

The idea to keep. For particles accelerated through the same voltage, λ = h/√2mqV. If the charges are equal, λ ∝ 1/√m, so the square root of the mass ratio always appears.

Answer from the NTA final answer key. Solution by PrepWiser.