An α-particle of kinetic energy 4.0 MeV moves head-on towards a tin nucleus (\(Z=50\)). The distance of closest approach of the α-particle to the centre of the nucleus is (Take \(1/4\pi\varepsilon_0=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\), \(e=1.6\times10^{-19}\ \mathrm{C}\) and \(1\ \mathrm{MeV}=1.6\times10^{-13}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(3.6\times10^{-14}\ \mathrm{m}\)
At the closest point, the α-particle stops for a moment. All its kinetic energy has then changed into electric potential energy. The α-particle carries a charge \( 2e \), and the tin nucleus carries \( Ze \).
\( \begin{aligned} K &= 4.0\times1.6\times10^{-13}\ \mathrm{J} \\ &= 6.4\times10^{-13}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} d &= \frac{2Ze^{2}}{4\pi\varepsilon_0K} \\ &= \frac{2\times50\times9\times10^{9}\times(1.6\times10^{-19})^{2}}{6.4\times10^{-13}} \\ &= \frac{2.30\times10^{-26}}{6.4\times10^{-13}} \\ &= 3.6\times10^{-14}\ \mathrm{m} \end{aligned} \)
Therefore, the α-particle turns back about 36 fm from the centre of the tin nucleus. Here 1 fm (fermi) is \( 10^{-15}\ \mathrm{m} \).
NCERT (2026-27):
Example 12.2 – “… Thus the distance of closest approach d is given by”
\( \displaystyle d=\frac{2Ze^{2}}{4\pi\varepsilon_0K} \)