In the Geiger-Marsden experiment, the distance of closest approach of a 5.5 MeV \(\alpha\)-particle to a gold nucleus \({}^{197}_{79}\mathrm{Au}\) is about \(4.0\times10^{-14}\ \mathrm{m}\). This distance is about how many times the radius of the gold nucleus?
(Take \(R_0=1.2\ \mathrm{fm}\) and \(197^{1/3}\approx5.8\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6
The radius of a nucleus depends on its mass number as \( R=R_0A^{1/3} \). For gold, \( A=197 \). We find this radius and compare the distance of closest approach with it.
\( \begin{aligned} R &= R_0A^{1/3} \\ &= 1.2\times5.8\ \mathrm{fm} \\ &\approx 7.0\ \mathrm{fm} \end{aligned} \)
The distance of closest approach is \( 4.0\times10^{-14}\ \mathrm{m}=40\ \mathrm{fm} \).
\( \begin{aligned} \frac{d}{R} &= \frac{40\ \mathrm{fm}}{7.0\ \mathrm{fm}} \\ &\approx 5.7 \\ &\approx 6 \end{aligned} \)
Hence, the α-particle turns back at about 6 times the radius of the gold nucleus. At this distance only the Coulomb repulsion acts on it.
NCERT (2026-27):
Size of the Nucleus – “It has been found that a nucleus of mass number A has a radius”
\( \displaystyle R=R_0A^{1/3} \)\( \qquad (13.5) \)
“where R0 = 1.2 × 10–15 m (=1.2 fm; 1 fm = 10–15 m). …”