Q.The Q value of a nuclear reaction $A + b \rightarrow C + d$ is defined by $Q = [\,m_A + m_b - m_C - m_d\,]c^2$ where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.

Concept: Nuclear Reaction Balancing — the $Q$-value tells us whether energy is released (exothermic, $Q>0$) or absorbed (endothermic, $Q<0$).

Step 1 – Reaction (i):

$^{1}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{2}_{1}\text{H} + {}^{2}_{1}\text{H}$

Mass of $^{1}_{1}\text{H}$ is $1.007825\ \text{u}$ (proton mass, standard value).

$Q = [1.007825 + 3.016049 - 2 \times 2.014102]\,c^2$

$= [4.023874 - 4.028204]\,c^2 = (-0.004330\ \text{u})\,c^2$

Using $1\ \text{u} = 931.5\ \text{MeV}/c^2$,

$Q = -0.004330 \times 931.5 \approx -4.03\ \text{MeV}$.

Step 2 – Reaction (ii):

$^{12}_{6}\text{C} + {}^{12}_{6}\text{C} \rightarrow {}^{20}_{10}\text{Ne} + {}^{4}_{2}\text{He}$ …

For each reaction, $Q = \big[\,\sum m_{\text{reactants}} - \sum m_{\text{products}}\,\big]\times 931.5\ \text{MeV}$. A positive $Q$ means energy is released (exothermic); a negative $Q$ means energy is absorbed (endothermic). Besides the masses listed in the question, the two standard atomic masses $m(^{1}_{1}\text{H}) = 1.007825\ \text{u}$ and $m(^{4}_{2}\text{He}) = 4.002603\ \text{u}$ are used.

(i) $^{1}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{2}_{1}\text{H} + {}^{2}_{1}\text{H}$

Since $Q<0$, the reaction is endothermic.

(ii) $^{12}_{6}\text{C} + {}^{12}_{6}\text{C} \rightarrow {}^{20}_{10}\text{Ne} + {}^{4}_{2}\text{He}$

$\Delta m = 2\,m(^{12}_{6}\text{C}) - \big[\,m(^{20}_{10}\text{Ne}) + m(^{4}_{2}\text{He})\,\big]$ …

Method: Mass-Energy Balance (Q-value calculation using atomic masses)

The Q-value tells you how much energy is released or absorbed in a nuclear reaction. The sign of Q decides whether the reaction is exothermic (energy released, Q > 0) or endothermic (energy absorbed, Q < 0).

Steps for any reaction:

1. Write the reaction in the form $A + b \rightarrow C + d$.
2. Identify the masses of all reactants and products from the given data. If a mass is not directly given, check if it can be inferred (here all are given).
3. Compute the mass difference: $\Delta m = (m_A + m_b) - (m_C + m_d)$.
4. Multiply by $c^2$ to get Q. In atomic mass units, $1\ \text{u} \cdot c^2 = 931.5\ \text{MeV}$.
5. If Q > 0, the reaction is exothermic. If Q < 0, it is endothermic.

(i) $^{1}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{2}_{1}\text{H} + {}^{2}_{1}\text{H}$

Masses:

• $m(^{1}_{1}\text{H})$ is not directly given. But note: $^{1}_{1}\text{H}$ is a proton. Its mass is not the same as the atomic mass of hydrogen-1 (which includes one electron). However, in nuclear reaction balancing using atomic masses, the electron masses cancel if the number of electrons is the same on both sides. Let’s check: Left side: 1 electron (from H-1) + 1 electron (from H-3) = 2 electrons. Right side: 1 electron (from H-2) + 1 electron (from H-2) = 2 electrons. So electron masses cancel. We can use the given atomic masses directly.

But we still need $m(^{1}_{1}\text{H})$. It is the atomic mass of hydrogen-1, which is $1.007825\ \text{u}$ (standard value). Since the problem does not provide it, we must use the standard value. (If the problem expected you to compute without it, they would have given it — but here it's missing. We proceed with the known value.)

$m(^{1}_{1}\text{H}) = 1.007825\ \text{u}$

$m(^{3}_{1}\text{H}) = 3.016049\ \text{u}$

$m(^{2}_{1}\text{H}) = 2.014102\ \text{u}$ (each)

Mass of reactants: $1.007825 + 3.016049 = 4.023874\ \text{u}$

Mass of products: $2 \times 2.014102 = 4.028204\ \text{u}$

Mass difference: $\Delta m = 4.023874 - 4.028204 = -0.004330\ \text{u}$

Q = $\Delta m \cdot c^2 = -0.004330 \times 931.5\ \text{MeV} \approx -4.03\ \text{MeV}$

Since Q < 0, the reaction is endothermic. …

Common Mistakes in Nuclear Reaction Q-Value Problems

Mistake 1: Using nuclear masses when only atomic masses are given

The formula uses nuclear masses, but the data gives atomic masses (which include electrons). For reactions where the number of electrons is conserved, atomic masses can be used directly — but students often forget to check this.

How to avoid: Count the electrons on each side. In reaction (i):

Left side: 1 electron (from $^1_1\text{H}$) + 1 electron (from $^3_1\text{H}$) = 2 electrons

Right side: 1 electron (from $^2_1\text{H}$) + 1 electron (from $^2_1\text{H}$) = 2 electrons

Electron count is balanced, so atomic masses work directly.

For reaction (ii):

Left: 6 + 6 = 12 electrons

Right: 10 + 2 = 12 electrons

Again balanced. So no correction needed here — but always verify.

Mistake 2: Forgetting to convert atomic mass units (u) to energy

The Q-value formula gives mass difference in u. To get energy in MeV, you must multiply by $931.5\ \text{MeV}/c^2$ (or equivalently, $c^2 = 931.5\ \text{MeV/u}$).

How to avoid: Always write the conversion explicitly:

Mistake 3: Sign error in the Q-value formula

Students sometimes write $Q = [m_C + m_d - m_A - m_b]c^2$ (reversed sign). This flips exothermic/endothermic classification.

How to avoid: Remember:

$Q > 0$ means exothermic (energy released) — the products have less mass than the reactants.

$Q < 0$ means endothermic (energy absorbed) — products have more mass.

The given formula $Q = [m_A + m_b - m_C - m_d]c^2$ is correct: reactants minus products.

Mistake 4: Arithmetic errors with precise mass values

The masses are given to 6 decimal places. A small subtraction error changes the sign or magnitude significantly.

How to avoid: Work systematically. For reaction (i):

Reactants: $m(^1_1\text{H}) + m(^3_1\text{H})$

We need $m(^1_1\text{H})$ — it's not given directly! This is a trap.

Using standard values:

$m(^1_1\text{H}) = 1.007825\ \text{u}$

$m(^3_1\text{H}) = 3.016049\ \text{u}$

Sum = $4.023874\ \text{u}$ …