Question

An $\alpha$ - particle & a proton are projected towards the same nucleus. Calculate the ratio of the distance of their closest approach if they are projected with the same kinetic energy.

Answer 1

$m_{\propto }=4amu$ $mp=1amu$

$\frac{K.{ϵ}_{\propto }}{K.{ϵ}_{\propto }}=\frac{1/2m_{\propto }{V}_{\propto }^2}{1/2m_p{V}_p^2}=1$

$\Rightarrow 4v_{\propto }^2=V{p}^2\Rightarrow Vp=2V_{\propto }$

we know that,

$V^2=u^2+2aS$

$V_p^2-u{p}^2=2aSp$

$V_{\propto }^2-u_{\propto }^2=2a{S}_{\propto }$

Initial velocities $up=u_{\propto }=0$

$u\frac{V_p^2}{V_{\propto }^2}=\frac{S_P}{S_{\propto }}\Rightarrow Sp:S_{\propto }=4:1$