Q.In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a $7.7\ \text{MeV}$ $\alpha$-particle before it comes momentarily to rest and reverses its direction?

The key idea is that at the distance of closest approach, the initial kinetic energy of the alpha particle is completely converted into electrostatic potential energy due to repulsion from the nucleus.

1. At the turning point, kinetic energy is zero. By energy conservation:

For gold ($Z=79$), the product $Ze$ is the nuclear charge.

2. Solve for the distance of closest approach $r_0$:

3. Substitute values. The constant $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2$, $e = 1.6 \times 10^{-19}\ \text{C}$, and $K_{\text{initial}} = 7.7\ \text{MeV} = 7.7 \times 1.6 \times 10^{-13}\ \text{J}$. …

The Geiger-Marsden experiment (Rutherford’s gold foil experiment) showed that the atom has a tiny, dense, positively charged nucleus. When an alpha particle heads straight toward the nucleus, it slows down as it climbs the Coulomb repulsion hill. At the point of closest approach, its kinetic energy has been completely converted into electrostatic potential energy — it comes momentarily to rest before being repelled back.

This is a pure energy conservation problem. No need to solve equations of motion; just set the initial kinetic energy equal to the potential energy at the turning point.

1. Write the energy conservation equation

The alpha particle starts far away (where potential energy is effectively zero) with kinetic energy $K = 7.7\ \text{MeV}$. At the distance of closest approach $r_0$, its speed is zero, so all energy is electrostatic potential energy:

Here:

• $Ze$ is the charge of the gold nucleus ($Z = 79$ for gold)
• $2e$ is the charge of the alpha particle
• $e = 1.6 \times 10^{-19}\ \text{C}$

2. Solve for $r_0$

3. Plug in the numbers

First, convert the kinetic energy to joules:

The Coulomb constant is:

Now:

Compute step by step:

• $2 \times 79 = 158$
• $(1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38}$
• Numerator: $158 \times 2.56 \times 10^{-38} = 4.0448 \times 10^{-36}$ …

Method: Conservation of Energy (Turning-Point Analysis)

This is a pure energy-conservation problem. The alpha particle approaches the nucleus head-on, slows down as its kinetic energy converts to electrostatic potential energy, and stops exactly at the distance of closest approach — the turning point.

Step 1: Identify the physical principle

At the moment of closest approach, the alpha particle's speed is zero. All its initial kinetic energy has been converted into electric potential energy between the alpha particle (charge $+2e$) and the gold nucleus (charge $+Ze$, where $Z=79$ for gold).

Step 2: Write the energy conservation equation

Initial kinetic energy $K_i$ = Final potential energy $U_f$ at distance $r_0$:

Step 3: Solve for $r_0$

Step 4: Plug in the numbers

• $K_i = 7.7\ \text{MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19}\ \text{J} = 1.232 \times 10^{-12}\ \text{J}$
• $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N·m}^2/\text{C}^2$
• $e = 1.6 \times 10^{-19}\ \text{C}$
• $Z = 79$

Step 5: Calculate step by step …

The most common mistakes on this question come from rushing through the physics and misapplying the energy conservation equation. Let me walk through each error and how to fix it.

Mistake 1: Forgetting that the alpha particle has two protons

Students often treat the alpha particle as a single charge $+e$ instead of $+2e$. The nucleus of gold has $Z = 79$, so the product $Z_1 Z_2$ becomes $2 \times 79 = 158$, not $1 \times 79$.

How to avoid: Always write down the atomic numbers explicitly before plugging in. Alpha particle: $Z_1 = 2$. Gold nucleus: $Z_2 = 79$. Then $Z_1 Z_2 = 158$.

Mistake 2: Using kinetic energy in eV instead of joules

The given energy is $7.7\ \text{MeV}$. If you plug $7.7 \times 10^6$ directly into the formula without converting to joules, you'll be off by a factor of $1.6 \times 10^{-19}$.

How to avoid: Convert MeV to joules immediately:

Mistake 3: Using the wrong formula — mixing up closest approach with impact parameter

The distance of closest approach $r_0$ for a head-on collision (where the alpha particle comes momentarily to rest) comes from equating initial kinetic energy to electrostatic potential energy at the turning point:

Some students mistakenly use the formula for impact parameter $b$ (which involves scattering angle) or the Rutherford scattering cross-section formula.

How to avoid: Remember the physical picture: "momentarily to rest" means all kinetic energy has converted to electrostatic potential energy. That's a straight energy conservation statement — no angles, no impact parameter.

Mistake 4: Forgetting the factor of $2$ in the denominator of Coulomb's constant

The constant $\frac{1}{4\pi\varepsilon_0}$ is $9 \times 10^9\ \text{N m}^2/\text{C}^2$. Some students use $k = 9 \times 10^9$ but then forget the $4\pi$ is already absorbed — they double-count it.

How to avoid: Use the standard value directly:

No further division by $4\pi$ is needed.

Mistake 5: Arithmetic errors in the final calculation

Even with the correct setup, the numbers are large and small simultaneously — $10^{-12}$ J, $10^{-19}$ C, $10^9$ constant. It's easy to misplace a power of 10.

How to avoid: Work systematically with powers of 10. Write the calculation step by step: …