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. (i) $^{1}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{2}_{1}\text{H} + {}^{2}_{1}\text{H}$ (ii) $^{12}_{6}\text{C} + {}^{12}_{6}\text{C} \rightarrow {}^{20}_{10}\text{Ne} + {}^{4}_{2}\text{He}$ Atomic masses are given to be $m\left(^{2}_{1}\text{H}\right) = 2.014102\ \text{u}$, $m\left(^{3}_{1}\text{H}\right) = 3.016049\ \text{u}$, $m\left(^{12}_{6}\text{C}\right) = 12.000000\ \text{u}$, $m\left(^{20}_{10}\text{Ne}\right) = 19.992439\ \text{u}$.

The Q value of a nuclear reaction A + b arrow C + – Class 12 Physics

Exercises · 13.15

A + b \rightarrow C + d

(i)

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

(ii)

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

Atomic masses are given to be

m\left(^{2}_{1}\text{H}\right) = 2.014102\ \text{u}

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

m\left(^{12}_{6}\text{C}\right) = 12.000000\ \text{u}

m\left(^{20}_{10}\text{Ne}\right) = 19.992439\ \text{u}

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Concept understanding — Nuclear Reaction Balancing

Nuclear Reaction Balancing: The Intuition

Think of a nuclear reaction like a game of atomic Lego. You start with a certain set of blocks (the reactants), and after the reaction, you end up with a different set of blocks (the products). The fundamental rule is: you cannot lose or gain any Lego pieces. You can rearrange them, break some apart, or fuse them together, but the total number of each type of piece must stay the same.

In the atomic world, the "pieces" are:

Protons (positive charge, found in the nucleus)
Neutrons (neutral charge, also in the nucleus)
Energy (which can appear or disappear, but that's a separate story)

The nucleus of an atom is made of protons and neutrons. When a nuclear reaction happens, the nuclei change. But the total number of protons and the total number of neutrons must be conserved — they cannot be created or destroyed.

Note

This is different from chemical reactions, where atoms themselves are conserved. In nuclear reactions, atoms can change into different elements, but the nucleons (protons + neutrons) are conserved.

The Precise Statement

A nuclear reaction is balanced when two quantities are equal on both sides of the reaction arrow:

Mass number (A) — the total number of nucleons (protons + neutrons). This is the superscript number.
Atomic number (Z) — the total number of protons. This is the subscript number.

For any nuclear reaction:

\text{Reactant}_1 + \text{Reactant}_2 \rightarrow \text{Product}_1 + \text{Product}_2 + \dots

The balancing conditions are:

\sum A_{\text{reactants}} = \sum A_{\text{products}}

\sum Z_{\text{reactants}} = \sum Z_{\text{products}}

\text{Total mass number (A) on left} = \text{Total mass number (A) on right}

How to Write a Nuclear Equation

Every nuclear particle is written as:

{}^{A}_{Z}X

Where:

$X$ = chemical symbol of the element
$A$ = mass number (top left)
$Z$ = atomic number (bottom left)

Common particles you'll encounter:

Particle	Symbol	$A$	$Z$
Alpha particle	$\alpha$ or ${}^{4}_{2}\text{He}$	4	2
Beta particle	$\beta^-$ or ${}^{0}_{-1}e$	0	-1
Gamma ray	$\gamma$ or ${}^{0}_{0}\gamma$	0	0
Neutron	$n$ or ${}^{1}_{0}n$	1	0
Proton	$p$ or ${}^{1}_{1}p$	1	1
Positron	$\beta^+$ or ${}^{0}_{+1}e$	0	+1

Watch out

A common mistake: forgetting that beta particles have $Z = -1$ (for $\beta^-$ ) or $Z = +1$ (for $\beta^+$ ). This is because a neutron turns into a proton (or vice versa), and the beta particle carries away the "missing" charge.

Worked Example

Problem: Balance the following alpha decay reaction:

{}^{238}_{92}\text{U} \rightarrow {}^{234}_{90}\text{Th} + \text{?}

Step 1: Identify what's missing. We have an unknown particle on the right.

Step 2: Balance mass numbers (A).

Left: $A = 238$

Right: $A = 234 + A_{\text{unknown}}$

So $238 = 234 + A_{\text{unknown}} \implies A_{\text{unknown}} = 4$

Step 3: Balance atomic numbers (Z).

Left: $Z = 92$

Right: $Z = 90 + Z_{\text{unknown}}$ …

Why this formula?

Why Nuclear Reaction Balancing Works

Nuclear reaction balancing rests on a single, non-negotiable principle: conservation laws are absolute. In every nuclear reaction — whether natural decay, artificial transmutation, or fission/fusion — two quantities never change:

Total mass number ( $A$ ) — the sum of protons + neutrons
Total atomic number ( $Z$ ) — the sum of protons

These aren't arbitrary rules. They follow from deeper physics: baryon number conservation (protons and neutrons are baryons, and their total count is fixed) and charge conservation (electric charge cannot be created or destroyed). A nuclear reaction is just a rearrangement of nucleons; the number of nucleons stays constant, and the total charge stays constant.

_{Z_1}^{A_1}X + _{Z_2}^{A_2}Y \rightarrow _{Z_3}^{A_3}W + _{Z_4}^{A_4}Z

The Reasoning Behind Each Conservation Law

Mass number conservation ( $A$ conserved):

A nucleon (proton or neutron) can change identity — a neutron can beta-decay into a proton, or a proton can capture an electron and become a neutron — but it cannot vanish or appear from nothing. The total count of nucleons before the reaction equals the total count after. This is why, for example, in alpha decay:

_{92}^{238}\text{U} \rightarrow _{90}^{234}\text{Th} + _{2}^{4}\text{He}

The left side has $A = 238$ ; the right side has $234 + 4 = 238$ . The alpha particle carries away exactly 4 nucleons.

Atomic number conservation ( $Z$ conserved):

Charge is strictly conserved. The total positive charge (proton count) before equals the total after. In the same alpha decay, $Z$ goes from 92 to $90 + 2 = 92$ . If charge weren't conserved, atoms would spontaneously change their chemical identity — which never happens in a closed system.

Watch out

A common mistake is to think mass number conservation means mass is conserved. It does not. Mass-energy is conserved, but the rest mass can change (and usually does, releasing energy). The mass number $A$ is a count of nucleons, not a measure of mass in kilograms.

How to Apply It: A Worked Example

Suppose you see: $_{92}^{235}\text{U} + _{0}^{1}n \rightarrow _{56}^{141}\text{Ba} + _{?}^{?}\text{Kr} + 3\, _{0}^{1}n$

You know the total $A$ on the left: $235 + 1 = 236$ .

On the right, you have $141 + A_{\text{Kr}} + 3(1) = 144 + A_{\text{Kr}}$ . …

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$Q = -4.03\ \text{MeV}$ — endothermic;
$Q = +4.62\ \text{MeV}$ — exothermic.

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

\Delta m = \big[\,m(^{1}_{1}\text{H}) + m(^{3}_{1}\text{H})\,\big] - 2\,m(^{2}_{1}\text{H})

\Delta m = (1.007825 + 3.016049) - 2(2.014102) = 4.023874 - 4.028204 = -0.004330\ \text{u}.

Q = (-0.004330)(931.5) = -4.03\ \text{MeV}.

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

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

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

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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:

Write the reaction in the form $A + b \rightarrow C + d$ .
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).
Compute the mass difference: $\Delta m = (m_A + m_b) - (m_C + m_d)$ .
Multiply by $c^2$ to get Q. In atomic mass units, $1\ \text{u} \cdot c^2 = 931.5\ \text{MeV}$ .
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. …

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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.

Watch out

If the electron count is not balanced, you must subtract the electron masses from the atomic masses to get nuclear masses, or add/subtract the appropriate number of electron masses to the Q-value.

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:

Q = (\Delta m \text{ in u}) \times 931.5\ \text{MeV/u}

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.

Note

The mass of $^1_1\text{H}$ (protium) is approximately $1.007825\ \text{u}$ (standard value). But since it's not in the given data, you must use the standard value or note that the problem expects you to know it.

Using standard values:

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

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

Sum = $4.023874\ \text{u}$ …

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