In Rutherford's experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to

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Q. In Rutherford's experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

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Concept understanding — Rutherford Scattering

Rutherford Scattering: From Intuition to Precision

Imagine you're in a dark room, and someone tells you there's a heavy object hidden somewhere. You can't see it, but you have a handful of marbles. If you roll marbles toward where you think the object might be, what happens? Most marbles roll straight through. But occasionally, one hits something hard and bounces back at a sharp angle. By studying how the marbles scatter — how many bounce back, and at what angles — you can figure out where the heavy object is, how big it is, and even how hard it is.

That's exactly what Rutherford did — but with alpha particles (tiny, fast, positively charged "marbles") and a thin gold foil (the "dark room").

The Intuition: Why Do Particles Scatter?

Before Rutherford, the atom was thought to be a "plum pudding" — a diffuse, positively charged blob with electrons scattered inside like raisins. If that were true, alpha particles fired at a gold foil should pass through almost undeflected, like bullets through jelly. The positive charge was spread out, so there was no concentrated "hard spot" to push them away.

But in 1909, Geiger and Marsden (working under Rutherford) found something shocking: a tiny fraction of alpha particles bounced back — some even at angles greater than 90°. Rutherford later said, "It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

The only explanation: the atom's positive charge and most of its mass are concentrated in a tiny, dense nucleus. When an alpha particle gets close to this nucleus, the strong electrostatic repulsion (both are positively charged) violently deflects it.

The Precise Statement

Rutherford scattering describes the elastic scattering of a charged particle (like an alpha particle) by a stationary, point-like nucleus, due to the Coulomb force (electrostatic repulsion). The key result is a formula for the number of particles scattered into a given direction.

N(\theta) = \frac{N_i n t Z_1^2 Z_2^2 e^4}{16 (4\pi\varepsilon_0)^2 r^2 E_k^2} \cdot \frac{1}{\sin^4(\theta/2)}

The most important part is the $\mathbf{1/\sin^4(\theta/2)}$ dependence. This tells you:

Most particles scatter at very small angles (they pass far from the nucleus and feel only a weak push).
Very few scatter at large angles (they must come extremely close to the nucleus).
The number of particles scattered at a given angle drops sharply as the angle increases.

What This Tells Us

The nucleus is tiny. For gold, the nuclear radius is about $10^{-14}$ m, while the atom's radius is about $10^{-10}$ m — the nucleus is 10,000 times smaller. That's why only 1 in 8,000 alpha particles bounced back.
The nucleus is positively charged. The repulsion is electrostatic, so the projectile and nucleus must have the same sign of charge.
The force is Coulombic. The $\sin^4(\theta/2)$ dependence is a direct consequence of the inverse-square law ( $F \propto 1/r^2$ ). If the force were different (e.g., short-range nuclear force), the angular distribution would change.
The scattering is elastic. The alpha particle loses no kinetic energy — it just changes direction. The nucleus recoils slightly, but the total kinetic energy is conserved.

Watch out

A common mistake: thinking Rutherford scattering applies to any particle hitting a nucleus. It only works when the projectile's energy is low enough that it never gets close enough to feel the strong nuclear force. At higher energies, the particle can penetrate the nucleus, and the scattering becomes inelastic (nuclear reactions occur). The formula above breaks down in that regime.

The Key Assumptions (Why the Formula Works)

Rutherford made several simplifying assumptions that hold for his experiment:

Single scattering — each alpha particle interacts with only one nucleus (the foil is thin enough).
Point-like nucleus — the nucleus is much smaller than the distance of closest approach.
Coulomb force only — no other forces (gravity, nuclear) are significant.
Target nucleus is stationary — it's much heavier than the alpha particle, so its recoil is negligible.
Elastic collision — no energy is lost to radiation or excitation. …

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Using a light-atom foil (e.g., aluminium) instead of a heavy one (e.g., gold) in Rutherford’s experiment reduces the number of large-angle scatterings and the fraction of α-particles that bounce back, because the lighter nucleus imparts less Coulomb repulsion and is itself pushed back more — the scattering pattern shifts toward smaller angles.

Rutherford’s gold-foil experiment revealed that atoms have a tiny, dense, positively charged nucleus. The key observation was that a small fraction of α-particles (about 1 in 8000 for gold) were scattered through angles greater than 90°, and some even bounced back. This happens because the α-particle and the gold nucleus repel each other strongly due to their positive charges. The closer the α-particle gets to the nucleus, the larger the scattering angle.

Now, if you replace the gold foil with a foil made of a light element like aluminium, the physics changes in a fundamental way. The crucial difference is the mass of the nucleus. Gold has a mass number of about 197, while aluminium is about 27. An α-particle has a mass number of 4. So the α-particle is much lighter than a gold nucleus, but it is comparable in mass to an aluminium nucleus.

Here’s why that matters, step by step.

The centre-of-mass effect. In Rutherford’s original derivation, he assumed the target nucleus is infinitely heavy — it stays fixed. That’s an excellent approximation for gold: the α-particle (mass 4) hits a nucleus of mass 197, so the nucleus barely recoils. But for aluminium (mass 27), the nucleus is only about 7 times heavier than the α-particle. The nucleus will recoil significantly, carrying away some of the incident kinetic energy. This reduces the energy available for the α-particle to overcome the Coulomb barrier and get close to the nucleus.
The distance of closest approach. For a head-on collision, the distance of closest approach $r_0$ is given by equating the initial kinetic energy $K$ to the electrostatic potential energy at the turning point:

K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}

where $Z$ is the atomic number (79 for gold, 13 for aluminium). So $r_0 \propto Z/K$ . But here’s the catch: in a collision with a light nucleus, the kinetic energy in the centre-of-mass frame is less than the lab-frame kinetic energy because the nucleus recoils. The effective kinetic energy available for the Coulomb repulsion is

K_{\text{cm}} = \frac{m_{\text{Al}}}{m_{\text{Al}} + m_\alpha} K_{\text{lab}}

For gold, this factor is nearly 1; for aluminium, it’s about $27/(27+4) \approx 0.87$ . So the α-particle cannot get as close to an aluminium nucleus as it can to a gold nucleus for the same incident energy.

Scattering probability and angle. The famous Rutherford scattering formula gives the number of particles scattered into a given solid angle:

\frac{dN}{d\Omega} = \left( \frac{Ze^2}{16\pi\epsilon_0 K} \right)^2 \frac{N_t}{\sin^4(\theta/2)}

Here $N_t$ is the number of target nuclei per unit area. Notice the $Z^2$ dependence. For gold ( $Z=79$ ) versus aluminium ( $Z=13$ ), the ratio is $(79/13)^2 \approx 37$ . That means, for the same foil thickness and incident energy, gold scatters about 37 times more α-particles into a given angle than aluminium does. But this formula assumes a fixed nucleus. For a light nucleus, the formula must be modified by replacing $K$ with $K_{\text{cm}}$ and the scattering angle $\theta$ in the lab frame is different from the centre-of-mass angle. The net effect is that large-angle scattering becomes much rarer.

The fraction of backscattered particles. The fraction of α-particles scattered through angles greater than 90° is roughly proportional to $Z^2 / K^2$ . For gold, this fraction is about $1.3 \times 10^{-4}$ (1 in 8000). For aluminium, using the same incident energy, the fraction drops by a factor of about $(79/13)^2 \approx 37$ , so it becomes roughly $3.5 \times 10^{-6}$ — that’s about 1 in 300,000. In practice, you would observe almost no α-particles bouncing back from an aluminium foil. …

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Q.In Rutherford's experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

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