Q.Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius $2.0\ \text{fm}$.)

The key idea is that the potential barrier height equals the Coulomb potential energy at the distance of closest approach — when the two deuterons just touch.

Reasoning:

1. For two identical deuterons (each with charge $+e$), the centre-to-centre distance at contact is twice the radius:

   $r = 2 \times 2.0\ \text{fm} = 4.0\ \text{fm}$.

2. The Coulomb potential energy at this separation is

3. Using $e = 1.6 \times 10^{-19}\ \text{C}$, $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2$, and $r = 4.0 \times 10^{-15}\ \text{m}$: …

The key insight here is that the "potential barrier" in nuclear fusion is the Coulomb repulsion that two positively charged nuclei must overcome before the strong nuclear force can bind them. For a head-on collision, the closest approach before the nuclear surfaces meet is when the centres are separated by the sum of their radii. At that instant, all the kinetic energy of approach has been converted into electrostatic potential energy — and that potential energy is the barrier height.

Let’s work through it step by step.

1. Understand the geometry of "just touching" Each deuteron is a hydrogen isotope nucleus (one proton, one neutron) with charge $+e$. The problem tells us to treat them as hard spheres of radius $r = 2.0\ \text{fm}$. When they just touch in a head-on collision, the centre-to-centre distance $d$ is:

This is the separation at which the Coulomb barrier is maximum — any closer and the strong force would begin to dominate, but we are finding the height of the barrier, i.e., the energy needed to reach this point.

2. The Coulomb potential energy formula The electrostatic potential energy of two point charges $q_1$ and $q_2$ separated by distance $d$ is:

Here, each deuteron has charge $q = +e = 1.602 \times 10^{-19}\ \text{C}$. So $q_1 q_2 = e^2$.

3. Plug in the numbers — but watch the units We want the answer in electronvolts (eV) because nuclear energies are typically expressed that way. The constant $\frac{1}{4\pi\varepsilon_0} = 8.987 \times 10^9\ \text{N·m}^2/\text{C}^2$. But a much cleaner route: use the known value $ke^2 = 1.44\ \text{MeV·fm}$, where $k = \frac{1}{4\pi\varepsilon_0}$. This is a standard nuclear physics shortcut. …

Method: Coulomb Barrier Height from Point-Charge Repulsion at Contact

This is a direct application of the electrostatic potential energy between two point charges at a separation equal to the sum of their radii.

Step 1 – Identify the physical picture.

For a head-on collision, the two deuterons approach until their surfaces just touch. At that instant, the distance between their centres is $r = 2R$, where $R$ is the radius of one deuteron. The Coulomb repulsion at this separation gives the height of the potential barrier — the minimum kinetic energy each deuteron must have (in the centre-of-mass frame) to overcome the barrier.

Step 2 – Write the Coulomb potential energy.

The potential energy of two point charges $q_1$ and $q_2$ separated by distance $r$ is

Each deuteron has charge $+e$, so $q_1 = q_2 = e$.

Step 3 – Substitute the given numbers.

Given $R = 2.0\ \text{fm} = 2.0 \times 10^{-15}\ \text{m}$, the centre-to-centre distance at contact is

The Coulomb constant is

and $e = 1.60 \times 10^{-19}\ \text{C}$.

Step 4 – Compute the barrier height in joules, then convert to MeV.

First, $e^2 = 2.56 \times 10^{-38}\ \text{C}^2$. Then

Convert to MeV using $1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$: …

Common Mistakes in Rutherford Scattering / Potential Barrier Problems

This question from nuclear physics tests your understanding of Coulomb repulsion at the nuclear scale. The key is recognising that "height of the potential barrier" means the electrostatic potential energy when the two nuclei are just touching — not the force, not the field, and not the potential at infinity.

Here are the mistakes students most often make, and how to avoid each.

Mistake 1: Using the wrong distance in Coulomb's law

Students often plug in the radius of one deuteron ($2.0\ \text{fm}$) as the separation distance $r$. But the two deuterons touch when their centres are separated by the sum of their radii — that is $2.0 + 2.0 = 4.0\ \text{fm}$.

How to avoid: Draw a quick sketch. Two spheres of equal radius just touching — the centre-to-centre distance is $2R$, not $R$. For this problem, $r = 4.0\ \text{fm}$.

Mistake 2: Confusing potential energy with potential

The question asks for the height of the potential barrier, which is the potential energy (in joules or electronvolts), not the electric potential (in volts). The formula for electrostatic potential energy of two point charges is:

Each deuteron has charge $q = +e = 1.6 \times 10^{-19}\ \text{C}$. So $q_1 q_2 = e^2$.

How to avoid: Read the phrase "height of the potential barrier" as "potential energy at the point of closest approach". Always check units: if the answer should be in MeV, you're computing energy.

Mistake 3: Forgetting to convert units properly

The radius is given in femtometres ($1\ \text{fm} = 10^{-15}\ \text{m}$). Students sometimes treat fm as $10^{-13}$ cm or forget the conversion entirely. Also, the final answer is expected in MeV, so you need to convert joules to electronvolts.

How to avoid: Write every conversion explicitly:

Then compute $U$ in joules, and divide by $1.6 \times 10^{-19}$ to get eV, then by $10^6$ to get MeV.

Mistake 4: Using the wrong value of $\frac{1}{4\pi\epsilon_0}$

The constant $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2$ is standard, but students sometimes use $8.99 \times 10^9$ and then round inconsistently. That's fine — but the real trap is forgetting that $e^2$ has units of $\text{C}^2$, so the product $k e^2$ gives $\text{N m}^2$, which simplifies to joules when divided by $r$ in metres.

A useful shortcut for nuclear problems: …