The ratio of de-Broglie wavelength of an α particle and a proton accelerated from rest by the same potential is 1/√m, the value of m is -

Solution

Physics sir · 4 steps

Write λ in terms of mass, charge and voltage, then let the same V cancel in the ratio.

1. Energy from the voltage. A charge q accelerated from rest through a potential V gains kinetic energy qV. Its momentum is p = √2mK = √2mqV.

   λ = h/p = h/√2mqV

2. Take the ratio. h, the factor 2 and V are the same for both particles, so λ_α/λₚ = √((mₚ qₚ)/(m_α q_α)).
3. Put in values. An α particle has m_α = 4mₚ and q_α = 2e, while the proton has charge e. So λ_α/λₚ = √((mₚ × e)/(4mₚ × 2e)) = 1/√8.
4. So. Matching with 1/√m gives m = 8, option (c).

The idea to keep. For particles accelerated through the same voltage, λ ∝ 1/√mq. Always bring in both the mass and the charge, because the voltage hands out energy qV.

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