Rutherford’s experiments used α-particles of energy about 5.5 MeV, which give a distance of closest approach to a gold nucleus of about \( 4.0\times10^{-14}\,\mathrm{m} \). What happens if α-particles of higher energy are used with the same gold foil?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The distance of closest approach will be smaller.
At the distance of closest approach \( d \) the α-particle stops for a moment. Its whole kinetic energy \( K \) has then become electrostatic potential energy in the Coulomb field of the gold nucleus (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 the same gold foil, \( d \) is inversely proportional to the kinetic energy of the α-particle.
Option (1) The distance of closest approach will be smaller. — Correct. A more energetic α-particle can do more work against the Coulomb repulsion, so it gets closer to the nucleus before it stops and turns back.
Option (2) The distance of closest approach will be the same. — Incorrect. The distance depends on \( K \). It stays the same only for α-particles of the same energy.
Option (3) The distance of closest approach will be larger. — Incorrect. A larger distance would need α-particles of lower energy, which are stopped farther from the nucleus.
Option (4) The α-particles will pass through the nucleus undeflected. — Incorrect. The α-particle still feels the Coulomb repulsion of the nucleus and is deflected. When it comes close enough, the short-range nuclear forces also act on it. They change the scattering further.
NCERT (2026-27):
Size of the Nucleus – “If we use α-particles of higher energies than 5.5 MeV, the distance of closest approach to the gold nucleus will be smaller and at some point the scattering will begin to be affected by the short range nuclear forces, and differ from Rutherford’s calculations.”