Q.$^{3}_{2}\text{He}$ and $^{3}_{1}\text{H}$ nuclei have the same mass number. Do they have the same binding energy?

No, they do not have the same binding energy. The key idea is that binding energy depends on the proton-neutron composition, not just the mass number.

Reasoning:

1. Both $^{3}_{2}\text{He}$ and $^{3}_{1}\text{H}$ have mass number $A = 3$, but different atomic numbers: $Z = 2$ for helium-3 and $Z = 1$ for tritium.
2. Binding energy is given by $B = (Z m_p + N m_n - m_{\text{nucleus}})c^2$. The difference in $Z$ changes the Coulomb repulsion energy among protons — more protons mean greater electrostatic repulsion, which reduces binding energy. …

Why binding energy varies with Z

Binding energy is the energy required to break a nucleus into its individual protons and neutrons. It comes from the strong nuclear force holding nucleons together, but this force is affected by two key factors:

1. Number of nucleons ($A$) — more nucleons generally mean more binding, but the relationship isn't linear.
2. Proton count ($Z$) — protons repel each other electrically, so extra protons reduce the net binding.

The semi-empirical mass formula (Bethe-Weizsäcker formula) captures this:

The key term here is the Coulomb term ($-a_c \frac{Z(Z-1)}{A^{1/3}}$). For fixed $A$, a larger $Z$ means more electrostatic repulsion, which reduces the binding energy.

Step-by-step comparison

1. Identify the nuclei

   $^{3}_{2}\text{He}$ has $Z=2$, $N=1$ (one neutron, two protons).

   $^{3}_{1}\text{H}$ (tritium) has $Z=1$, $N=2$ (two neutrons, one proton).

   Both have $A=3$.

2. Apply the Coulomb term

   For $^{3}_{2}\text{He}$: $Z=2$, so $Z(Z-1)=2$. The Coulomb term is $-a_c \cdot \frac{2}{3^{1/3}}$.

   For $^{3}_{1}\text{H}$: $Z=1$, so $Z(Z-1)=0$. The Coulomb term is zero.

   This alone tells us $^{3}_{2}\text{He}$ has less binding energy due to proton-proton repulsion.

3. Consider the asymmetry term

   The term $-a_a \frac{(A-2Z)^2}{A}$ penalizes imbalance between protons and neutrons.

   For $^{3}_{2}\text{He}$: $A-2Z = 3-4 = -1$, so $(A-2Z)^2 = 1$.

   For $^{3}_{1}\text{H}$: $A-2Z = 3-2 = 1$, also $(A-2Z)^2 = 1$.

   So the asymmetry term is identical for both — no difference here.

4. Check the pairing term

   The pairing term $\delta(A,Z)$ accounts for whether $Z$ and $N$ are even or odd.

   $^{3}_{2}\text{He}$: $Z=2$ (even), $N=1$ (odd) → an even-$Z$, odd-$N$ configuration, which gives a small positive $\delta$ (extra binding).

   $^{3}_{1}\text{H}$: $Z=1$ (odd), $N=2$ (even) → odd-even, also a small positive $\delta$. …

Method: Comparing Binding Energies of Isobars (Same A, Different Z)

Use this whenever two nuclei share the same mass number $A$ but differ in proton number $Z$, and you must decide whether/how their binding energies differ.

Steps

Step 1: Recognize that binding energy depends on BOTH $A$ and $Z$, not $A$ alone

Binding energy is not simply a function of how many nucleons there are — it also depends on how those nucleons are split between protons and neutrons, because protons repel each other electrically while neutrons don't.

Step 2: Recall the relevant physics via the semi-empirical mass formula

You don't need to compute every term numerically — the key term for comparing isobars is the Coulomb term, $-a_c\,\dfrac{Z(Z-1)}{A^{1/3}}$, since $A$ is fixed and only $Z$ differs.

Step 3: Compare the Coulomb term for the two nuclei …