An α particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are λ α and λₚ respectively. The ratio (λ ₚ)/(λ _α) is :

Solution

Physics sir · 4 steps

Both gain kinetic energy qV, so λ = h/√2mqV; compare mq for the two.

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

   λ = h/p = h/√2mqV

2. Take the ratio. The same V acts on both, so h, the 2 and V cancel: λₚ/λ_α = √((m_α q_α)/(mₚ qₚ)).
3. Put in numbers. An α particle has m_α = 4mₚ and q_α = 2e, while the proton has qₚ = e. So λₚ/λ_α = √(4 × 2) = √8 = 2√2 ≈ 2.83.
4. So. The ratio is about 2.8, which is option (b).

The idea to keep. For charges accelerated through the same voltage, λ ∝ 1/√mq. Heavier or more highly charged particles get the shorter wavelength.

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