A deuterium nucleus and an alpha particle approach a target nucleus separately in head-on position. Find the ratio of their distance of closest…

$\displaystyle \dfrac{1}{2} \mathrm{mv}^{2}=\dfrac{k \mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{d}}$

$\displaystyle \dfrac{\mathrm{d}_{\text{deuteron}}}{\mathrm{d}_{\alpha}}=\dfrac{\mathrm{q}_{\text{deuteron}}}{\mathrm{q}_{\alpha}} \times \dfrac{\mathrm{m}_{\alpha}}{\mathrm{m}_{\text{deuteron}}}=\dfrac{e}{2 e} \times \dfrac{4 m}{2 m}=1$

$\displaystyle \mathrm{d}_{\text{deuteron}}: \mathrm{d}_{\alpha}=1: 1$

$\displaystyle \dfrac{\mathrm{d}_{\text{deuteron}}}{\mathrm{d}_{\alpha}}=\dfrac{\mathrm{q}_{\text{deuteron}} \times \mathrm{m}_{\text{deuteron}}}{\mathrm{q}_{\alpha} \times \mathrm{m}_{\alpha}}=\dfrac{e \times 2 m}{2 e \times 4 m}=\dfrac{1}{4}$

$\displaystyle \mathrm{d}_{\text{deuteron}}: d_{\alpha}=1: 4$