An α particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

Solution

Physics sir · 4 steps

Same voltage means kinetic energy qV, and momentum is √2mK, so compare mq under a root.

1. Energy gained. A charge q accelerated from rest through a potential difference V gains kinetic energy K = qV.
2. Energy to momentum. Since K = p²/2m, the momentum is p = √2mK. Put in K = qV to get the key relation below.

   p = √2mqV ⇒ p_α/pₚ = √((m_α q_α)/(mₚ qₚ))

3. Put in values. An α particle has m_α = 4mₚ and q_α = 2e, while the proton has qₚ = e. So the ratio is √(4 × 2) = √8 = 2√2.
4. So. The momenta are in the ratio p_α : pₚ = 2√2 : 1, which is option (b).

The idea to keep. For anything accelerated through a voltage, p = √2mqV. Both mass and charge sit under the square root, so multiply their ratios first and take the root last.

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