An $\alpha$-particle of energy $7.7\,\text{MeV}$ is directed head-on at a stationary gold nucleus ($Z = 79$). At the distance of CLOSEST APPROACH, the $\alpha$-particle momentarily comes to rest. The distance of closest approach is closest to (Take $\dfrac{1}{4\pi\varepsilon_0} = 9\times 10^9\,\text{N\,m}^2/\text{C}^2$ and $e = 1.6\times 10^{-19}\,\text{C}$.)

1. At the closest approach, the $\alpha$-particle has zero kinetic energy, so its initial KE has been entirely converted into electrostatic potential energy. 2. Conservation of energy: $K = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{(2e)(Ze)}{d}$ (since $\alpha$ carries charge $+2e$ and the nucleus $+Ze$). 3. Solve for $d$: $d = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{2Ze^2}{K}$. 4. Convert K: $K = 7.7\,\text{MeV} = 7.7\times 10^{6}\times 1.6\times 10^{-19} = 1.232\times 10^{-12}\,\text{J}$. 5. Plug in: $d = \dfrac{(9\times 10^{9})(2\times 79\times (1.6\times 10^{-19})^2)}{1.232\times 10^{-12}}$ = $\dfrac{(9\times 10^{9})(4.05\times 10^{-36})}{1.232\times 10^{-12}}$ ≈ $\dfrac{3.64\times 10^{-26}}{1.232\times 10^{-12}}$ ≈ $2.96\times 10^{-14}\,\text{m}$, i.e. about $3\times 10^{-14}\,\text{m}$. 6. This is roughly the size of a heavy nucleus, consistent with the Rutherford experiment. Other options are off by orders of magnitude. _Source: NCERT Class 12 Physics Part 2, Ch 12, Example 12.2 (Geiger–Marsden, $7.7\,\text{MeV}$ alpha), p. 4._