Alpha particles are helium nuclei, consisting of 2 protons and 2 neutrons (\$^{4}_{2}\mathrm{He}\$). They are relatively heavy and carry a +2 charge.
Think of them as tiny “mini‑bombs” that can knock other particles around.
🔍 Observation: The existence of a tiny, dense, positively charged centre in the atom.
Using the Coulomb force between the positively charged alpha particle and the nucleus:
\$F = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r^2}\$
When an alpha particle comes close to the nucleus, the repulsive force can be strong enough to change its direction dramatically.
From the maximum scattering angle (\$\theta_{\text{max}}\$) we can estimate the size of the nucleus:
\$r{\text{nucleus}} \approx \frac{1}{4\pi\varepsilon0}\,\frac{(2e)(Ze)}{E{\alpha}}\tan\left(\frac{\theta{\text{max}}}{2}\right)\$
Result: The nucleus is tiny (≈10⁻¹⁵ m) compared to the whole atom (≈10⁻¹⁰ m).
Imagine a bowling ball (alpha particle) rolling toward a pinball machine (atom). Most of the time it just slides past the machine, but occasionally it hits a heavy pin (nucleus) and gets knocked back or deflected sharply.
Just like the pin is tiny compared to the whole machine, the nucleus is tiny compared to the atom.
Use the structure:
Remember to include the Coulomb force equation and the scattering angle formula where relevant.
Suppose an alpha particle with kinetic energy \$E_{\alpha}=5\,\text{MeV}\$ is scattered at an angle of \$90^{\circ}\$. Estimate the radius of the gold nucleus (\$Z=79\$). Use the approximation:
\$r{\text{nucleus}} \approx \frac{1}{4\pi\varepsilon0}\,\frac{(2e)(Ze)}{E_{\alpha}}\$
Show your calculation steps.