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 :

Solution

Physics sir · 3 steps

Write λ in terms of mass, charge and voltage, then let the ratio do the work.

1. Energy from the voltage. A charge q accelerated from rest through a potential V gains kinetic energy qV. Its momentum is p = √2mqV, so the de Broglie wavelength λ = h/p becomes

   λ = h/√2mqV ⇒ λₚ/λ_α = √((m_α q_α V_α)/(mₚ qₚ Vₚ))

2. Put in numbers. An α-particle has m_α = 4mₚ and q_α = 2e. With Vₚ = 2 V and V_α = 4 V, the ratio is √((4 × 2 × 4)/(1 × 1 × 2)) = √16 = 4.
3. So. λₚ : λ_α = 4 : 1, which is option (a).

The idea to keep. For any charged particle accelerated from rest, λ ∝ 1/√mqV. Mass, charge and voltage all sit under one square root, so put every ratio inside it before taking the root.

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