Q.Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below $14\ \text{K}$.) What results do you expect?

The outcome is decided by the mass of the target nucleus, not by its charge alone. A solid-hydrogen sheet presents single protons ($\approx1\ \text{u}$), which are lighter than the incoming $\alpha$-particle ($4\ \text{u}$).

• Large-angle and back-scattering in the gold-foil experiment occurred because the gold nucleus ($A\approx197$) is far heavier than the $\alpha$ and acts like a fixed wall.
• A proton is $4$ times lighter than the $\alpha$, so it recoils and carries off energy instead of turning the $\alpha$ around. A light target can never scatter a heavier projectile through a large angle.
• For an elastic collision of a heavier projectile ($m=4\ \text{u}$) with a lighter stationary target ($M=1\ \text{u}$), the projectile's deflection is bounded by

So the $\alpha$-particles emerge within about $14^\circ$ of their original direction, with no back-scattering — the experiment would not reveal a tiny massive nucleus.

1. Why the target mass is decisive. In the gold-foil experiment the large-angle (and back-) scattering arose because the gold nucleus ($A\approx197$) is roughly $50$ times heavier than the $\alpha$-particle and behaves essentially like a fixed scattering centre. A proton, by contrast, is $4$ times lighter than the $\alpha$: instead of turning the $\alpha$ around, it recoils and is knocked forward.

2. Kinematic limit for a heavy projectile on a light target. For an elastic collision of a projectile of mass $m=4\ \text{u}$ with a stationary target of mass $M=1\ \text{u}$ (with $m>M$), the projectile's scattering angle in the lab frame can never exceed

Hence every $\alpha$-particle emerges within about $14^\circ$ of its incident direction; none is deflected backwards.

3. Where the energy goes. Being light, the struck proton recoils and carries off a substantial share of the kinetic energy, while the $\alpha$-particle continues almost straight ahead, only slightly slowed and slightly deflected.

4. Consequence for the experiment. The dramatic large-angle scattering that led Rutherford to infer a tiny, massive, positively charged nucleus would be absent. Repeating the experiment with solid hydrogen would therefore fail to reveal the nuclear structure of the atom.

Method: Impact Parameter Analysis in Rutherford Scattering

This problem tests your understanding of how the impact parameter (the perpendicular distance between the alpha particle's initial path and the nucleus) determines the scattering angle — and how that depends on the nuclear charge of the target.

Step 1 — Recall the key relationship

For Rutherford scattering, the distance of closest approach $r_0$ (for a head-on collision) is given by:

where $Z$ is the atomic number of the target nucleus, $e$ is the electronic charge, and $E$ is the kinetic energy of the alpha particle.

The scattering angle $\theta$ is related to the impact parameter $b$ by:

Step 2 — Compare the target nuclei

The alpha particle itself has $Z=2$ and mass ~4 u.

Step 3 — Analyse the two key effects

Effect 1: Much weaker Coulomb repulsion

Since $Z$ for hydrogen is 79 times smaller than for gold, the Coulomb force at the same distance is 79 times weaker. For the same impact parameter, the scattering angle will be drastically smaller. Most alpha particles will pass through with very little deflection — the scattering pattern will be much more concentrated near $0^\circ$.

Effect 2: Recoil of the target nucleus

This is the critical difference. In gold, the nucleus is massive (~197 u) compared to the alpha particle (~4 u), so it barely moves — the scattering is essentially elastic with a fixed centre. But hydrogen has a nucleus of mass ~1 u, which is lighter than the alpha particle.

When a moving alpha particle collides with a stationary hydrogen nucleus, the hydrogen nucleus can recoil significantly. In fact, in a head-on collision, the alpha particle can transfer a large fraction of its energy to the proton.

Step 4 — Predict the expected results

1. Most alpha particles will pass through with very small deflection angles (much smaller than with gold), because the nuclear charge is tiny.

2. A small fraction of alpha particles that come very close to a hydrogen nucleus will experience a large-angle scattering, but the kinematics will be different — the hydrogen nucleus will recoil, and the alpha particle will lose significant energy.

3. Head-on collisions will result in the alpha particle transferring most of its kinetic energy to the proton, and the alpha particle may even be scattered backwards in the centre-of-mass frame, but with much lower energy than in the gold case.

4. The scattering pattern will not follow the simple Rutherford $1/\sin^4(\theta/2)$ law because the target recoil cannot be ignored.

Final answer: With a solid hydrogen target, most alpha particles pass through undeflected or with tiny angles. Very few are scattered through large angles, and those that are lose significant energy to the recoiling proton. The classical Rutherford scattering formula does not apply because the target nucleus is lighter than the projectile.

Common Mistakes in the Rutherford Scattering Distance Problem (Hydrogen vs Gold Foil)

This question from the NCERT/CBSE syllabus tests your understanding of how the scattering formula depends on the nuclear charge $Z$ and the target thickness. Students typically make four distinct errors here.

Mistake 1: Forgetting that hydrogen has $Z = 1$

The most common error is to treat hydrogen as having the same nuclear charge as gold ($Z = 79$). The distance of closest approach in Rutherford scattering is given by:

Since $r_0 \propto Z$, using hydrogen ($Z = 1$) instead of gold ($Z = 79$) means the distance of closest approach becomes 79 times smaller. Students often miss this factor entirely.

How to avoid: Always write down the proportionality $r_0 \propto Z$ before plugging numbers. Compare the $Z$ values explicitly: $Z_{\text{Au}} = 79$, $Z_{\text{H}} = 1$.

Mistake 2: Ignoring the target thickness effect

Students often compare only the nuclear charge and forget that the foil is a thin sheet of solid hydrogen. The number of target nuclei per unit area ($N_t$) matters because the scattering probability depends on it.

For a given foil thickness $t$ and number density $n$:

Solid hydrogen has a much lower density than gold. At 14 K, solid hydrogen has a density of about $0.086\ \text{g/cm}^3$, compared to gold's $19.3\ \text{g/cm}^3$. The number density of hydrogen nuclei is:

For gold:

Surprisingly, the number densities are comparable! But the scattering cross-section per nucleus is proportional to $Z^2$, so the overall scattering rate drops by a factor of $79^2 \approx 6240$ for the same thickness.

How to avoid: Always check both factors: nuclear charge $Z$ and target thickness $N_t$. Write the full proportionality:

Mistake 3: Claiming "no scattering" or "all alpha particles pass through"

Some students conclude that because hydrogen has a tiny nucleus, alpha particles will never be scattered. This is incorrect — scattering still occurs, but the angles are much smaller and the probability of large-angle scattering is drastically reduced.

The impact parameter $b$ for a given scattering angle $\theta$ is:

Since $b \propto Z$, for the same scattering angle, the impact parameter with hydrogen is 79 times smaller than with gold. This means only alpha particles that come extremely close to the hydrogen nucleus (within about $10^{-15}\ \text{m}$) will scatter at large angles.

How to avoid: State clearly: "Scattering still occurs, but the cross-section is smaller by a factor of $Z^2 = 6241$, so large-angle scattering events become extremely rare."

Mistake 4: Confusing the target with the projectile

A subtle error: some students think that because hydrogen is light, the alpha particle might knock the hydrogen nucleus out of the foil (like in a collision between billiard balls). While this is true for close collisions (the hydrogen nucleus can recoil), the question specifically asks about Rutherford scattering of the alpha particle itself.

The Rutherford formula assumes the target nucleus is stationary and heavy. For hydrogen, this assumption breaks down for very close collisions — the hydrogen nucleus recoils significantly. However, for the typical scattering experiment, the alpha particle is much heavier than the proton ($m_\alpha \approx 4 m_p$), so the correction is not drastic.

How to avoid: Mention that the hydrogen nucleus can recoil, but the dominant effect is still the $Z^2$ reduction in scattering cross-section.

Final Expected Answer

With a solid hydrogen foil:

• The distance of closest approach decreases by a factor of 79.
• The scattering probability at any given angle decreases by a factor of $79^2 \approx 6240$.
• Large-angle scattering becomes extremely improbable — most alpha particles pass through undeflected or with very small deflections.
• The foil must be kept at cryogenic temperatures (below 14 K) to remain solid.