A proton and an alpha particle having equal velocities approach a target nucleus. They come momentarily to rest and then reverse their directions. The ratio of the distance of closest approach of the proton to that of the alpha particle will be :

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YESPYQ · Accepted Answer

Correct answer: b) 2. $ \text{Given:} \quad q_p = +e,\, m_p; \quad q_\alpha = +2e,\, m_\alpha = 4m_p; \quad v_p = v_\alpha = v $ $ \text{At the point of closest approach:} \quad \frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{Z e q}{r_{\min}} $ $ r_{\min} = \frac{1}{4\pi\varepsilon_0} \frac{2 Z e q}{m v^2} $ $ \text{For proton:} \quad r_p = \frac{2 Z e^2}{4\pi\varepsilon_0\, m_p v^2} $ $ \text{For alpha particle:} \quad r_\alpha = \frac{2 Z (2e)e}{4\pi\varepsilon_0\, (4m_p) v^2} $ $ \frac{r_p}{r_\alpha} = \frac{\dfrac{e}{m_p}}{\dfrac{2e}{4m_p}} = \frac{1}{(2/4)} = 2 $ $ \boxed{\frac{r_p}{r_\alpha} = 2} $

A proton and an alpha particle having equal velocities approach a target nucleus. They come momentarily to rest and then…

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