Read the following passage and answer the questions.
In 1911, H. Geiger and E. Marsden did a scattering experiment suggested by Rutherford. A radioactive source of \({}^{214}_{83}\mathrm{Bi}\) gave out alpha-particles of energy 5.5 MeV. The alpha-particles passed through lead bricks and then fell on a gold foil. The foil was only \(2.1\times10^{-7}\ \mathrm{m}\) thick. A detector made of a zinc sulphide screen and a microscope was placed near the foil. The detector could be turned around the foil. The whole apparatus was kept in a vacuum chamber. An alpha-particle carries a positive charge \(2e\). A gold nucleus carries a positive charge \(Ze\), with \(Z = 79\) for gold. The graph below shows the number of scattered alpha-particles detected at different scattering angles.
An alpha-particle of the experiment moves straight towards a gold nucleus. What is its kinetic energy when it is \(1.0\times10^{-13}\ \mathrm{m}\) from the centre of the nucleus? (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 2
The correct answer is Option (2) → 3.2 MeV
The alpha-particle and the gold nucleus repel each other. As the alpha-particle comes closer, some of its kinetic energy changes into electric potential energy. The total mechanical energy of the two stays the same.
Given:
\( K_i=5.5\ \mathrm{MeV} \)
\( r=1.0\times10^{-13}\ \mathrm{m} \)
\( q_{\alpha}=2e \)
\( Z=79 \)
Potential energy at this point:
\( \begin{aligned} U &= \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r} \\ &= \frac{9\times10^{9}\times2\times79\times(1.6\times10^{-19})^{2}}{1.0\times10^{-13}} \\ &= 3.64\times10^{-13}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} U &= \frac{3.64\times10^{-13}}{1.6\times10^{-13}}\ \mathrm{MeV} \\ &\approx 2.3\ \mathrm{MeV} \end{aligned} \)
Kinetic energy at this point:
\( \begin{aligned} K &= K_i-U \\ &= 5.5-2.3 \\ &= 3.2\ \mathrm{MeV} \end{aligned} \)
From the above, \( K \approx 3.2\ \mathrm{MeV} \). Hence, the alpha-particle is still moving at this point. It has not yet reached its distance of closest approach.
NCERT (2026-27):
Example 12.2 – “Solution The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an α-particle and a gold nucleus is conserved. …”
“Let d be the centre-to-centre distance between the α-particle and the gold nucleus when the α-particle is at its stopping point. Then we can write the conservation of energy Ei = Ef as”
\( \displaystyle K=\frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{d}=\frac{2Ze^{2}}{4\pi\varepsilon_0 d} \)