An alpha particle of energy 5 MeV is scattered through \( 180^\circ \) by a fixed uranium nucleus. The distance of closest approach is of the order of
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 10^{-12}\,\mathrm{cm} \)
An alpha particle scattered through \( 180^\circ \) makes a head-on approach: it stops momentarily and turns back. At the distance of closest approach \( d \), all its kinetic energy \( K \) has changed into electric potential energy (conservation of energy). The alpha particle has charge \( 2e \), and the nucleus has charge \( Ze \).
\( \displaystyle K=\frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{d} \)
\( \displaystyle \Rightarrow\; d=\frac{2Ze^{2}}{4\pi\varepsilon_0K} \)
For uranium \( Z=92 \). Taking \( 1/4\pi\varepsilon_0=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \), \( e=1.6\times10^{-19}\,\mathrm{C} \) and \( 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J} \):
\( \begin{aligned} K &= 5\times1.6\times10^{-13}\,\mathrm{J} \\ &= 8\times10^{-13}\,\mathrm{J} \end{aligned} \)
\( \begin{aligned} d &= \frac{2\times92\times9\times10^{9}\times(1.6\times10^{-19})^{2}}{8\times10^{-13}}\,\mathrm{m} \\ &\approx 5.3\times10^{-14}\,\mathrm{m} \\ &= 5.3\times10^{-12}\,\mathrm{cm} \end{aligned} \)
Therefore, the distance of closest approach is of the order of \( 10^{-12}\,\mathrm{cm} \).