In a head-on collision between an alpha particle and a gold nucleus (Z = 79), the distance of closest approach is 41.3 fm. The energy of the alpha particle is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5.51 MeV
In a head-on collision the alpha particle stops momentarily at the distance of closest approach; there all its kinetic energy has become electrostatic potential energy of the alpha particle (charge 2e) and the gold nucleus (charge Ze).
\( \displaystyle K = \frac{1}{4\pi\varepsilon_0}\,\frac{(Ze)(2e)}{r_0} \)
\( Z = 79 \)
\( r_0 = 41.3\,\mathrm{fm} = 41.3\times10^{-15}\,\mathrm{m} \)
\( \displaystyle K = \frac{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \times 79 \times 2 \times (1.6\times10^{-19}\,\mathrm{C})^{2}}{41.3\times10^{-15}\,\mathrm{m}} \)
\( \displaystyle K \approx 8.81\times10^{-13}\,\mathrm{J} \)
Converting with \( 1\,\mathrm{MeV} = 1.6\times10^{-13}\,\mathrm{J} \):
\( \displaystyle K = \frac{8.81\times10^{-13}}{1.6\times10^{-13}}\,\mathrm{MeV} \approx 5.51\,\mathrm{MeV} \)
Shortcut: Use \( \frac{e^{2}}{4\pi\varepsilon_0} = 1.44\,\mathrm{MeV\,fm} \): \( K = \frac{1.44 \times 79 \times 2}{41.3} \) MeV \( \approx 5.51 \) MeV, with no conversion of units.