Question

An $\alpha$ -particle and a proton are accelerated through the same potential difference. The ratio of linear momenta acquired by the two is ;

• A. $\sqrt{4}:1$
• B. $\sqrt{2}:1$
• C. $1:\sqrt{2}$
• D. $\sqrt{8}:1$

Answer 1

Answer: (D)

$\sqrt{8}:1$

Answer 2

: The kinetic energy gained by a particle when acceleration through potential difference V is

}{orm^{2}v^{2} = 2mqV........... ⥂ (i)}$$ $\therefore$Momentum . $p = mv = \sqrt{2mqV}(u\sin g(i))$ Or $\frac{p_{\alpha}}{p_{p}} = \sqrt{\frac{2m_{\alpha}q_{\alpha}V}{2m_{p}q_{p}V}} = \sqrt{\frac{2 \times 4m_{p} \times 2e \times V}{2 \times m_{p} \times e \times V}} = \sqrt{8}:1$