Suppose, we think of fission of a $^{56}_{26}\text{Fe}$ nucleus into two equal fragments, $^{28}_{13}\text{Al}$. Is the fission energetically possible? Argue by working out $Q$ of the process. Given $m\left(^{56}_{26}\text{Fe}\right) = 55.93494\ \text{u}$ and $m\left(^{28}_{13}\text{Al}\right) = 27.98191\ \text{u}$.

> [!TLDR] > Fission of $^{56}_{26}\text{Fe}$ into two $^{28}_{13}\text{Al}$ nuclei is **not energetically possible** because the $Q$-value is negative ($Q \approx -0.02888\ \text{u} \approx -26.9\ \text{MeV}$), meaning energy must be supplied rather than released. ## The Core Idea: Why $Q$ Tells Us If a Reaction "Goes" Nuclear reactions, like chemical reactions, obey energy conservation. But in nuclear physics, mass *is* a form of energy — Einstein's $E = mc^2$. So when a nucleus splits (fission) or joins (fusion), we don't just track kinetic energy; we track the **mass defect**. The $Q$-value of a nuclear reaction is the **energy released** (or absorbed) during the process. It's defined as: $$Q = (\text{total mass of reactants} - \text{total mass of products}) \times c^2$$ If $Q > 0$, the reaction is **exothermic** — mass is converted to energy, and the reaction can happen spontaneously. If $Q [!WARNING] > A common mistake is to forget that the **number of nucleons** must balance on both sides. Here, $^{56}_{26}\text{Fe}$ has 56 nucleons; two $^{28}_{13}\text{Al}$ nuclei also have $2 \times 28 = 56$ nucleons. So the reaction is *possible* in terms of nucleon conservation — but energetics is another matter. ## Step-by-Step Calculation **1. Write the reaction clearly** We're considering: $$^{56}_{26}\text{Fe} \rightarrow \, ^{28}_{13}\text{Al} + \, ^{28}_{13}\text{Al}$$ That's one iron nucleus splitting into two identical aluminum nuclei. **2. Recall the $Q$-value formula** For a reaction $A \rightarrow B + C$: $$Q = [m(A) - m(B) - m(C)] \times c^2$$ If $Q$ comes out positive, mass has been lost and energy released. If negative, mass has been gained — meaning energy must be supplied. **3. Plug in the given masses** We have: - $m(^{56}_{26}\text{Fe}) = 55.93494\ \text{u}$ - $m(^{28}_{13}\text{Al}) = 27.98191\ \text{u}$ So: $$Q = [5

Suppose, we think of fission of a ⁵⁶₂₆Fe nucleus into two equal fragments, ²⁸₁₃Al. Is the fission energeticall

Exercises · 13.16

^{56}_{26}\text{Fe}

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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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Fission of $^{56}_{26}\text{Fe}$ into two $^{28}_{13}\text{Al}$ nuclei is not energetically possible because the $Q$ -value is negative ( $Q \approx -0.02888\ \text{u} \approx -26.9\ \text{MeV}$ ), meaning energy must be supplied rather than released.

The Core Idea: Why $Q$ Tells Us If a Reaction "Goes"

Nuclear reactions, like chemical reactions, obey energy conservation. But in nuclear physics, mass is a form of energy — Einstein's $E = mc^2$ . So when a nucleus splits (fission) or joins (fusion), we don't just track kinetic energy; we track the mass defect.

The $Q$ -value of a nuclear reaction is the energy released (or absorbed) during the process. It's defined as:

Q = (\text{total mass of reactants} - \text{total mass of products}) \times c^2

If $Q > 0$ , the reaction is exothermic — mass is converted to energy, and the reaction can happen spontaneously. If $Q < 0$ , the reaction is endothermic — you'd need to pump energy in to make it happen. That's the whole game here.

Watch out

A common mistake is to forget that the number of nucleons must balance on both sides. Here, $^{56}_{26}\text{Fe}$ has 56 nucleons; two $^{28}_{13}\text{Al}$ nuclei also have $2 \times 28 = 56$ nucleons. So the reaction is possible in terms of nucleon conservation — but energetics is another matter.

Step-by-Step Calculation

1. Write the reaction clearly

We're considering:

^{56}_{26}\text{Fe} \rightarrow \, ^{28}_{13}\text{Al} + \, ^{28}_{13}\text{Al}

That's one iron nucleus splitting into two identical aluminum nuclei.

2. Recall the $Q$ -value formula

For a reaction $A \rightarrow B + C$ :

Q = [m(A) - m(B) - m(C)] \times c^2

If $Q$ comes out positive, mass has been lost and energy released. If negative, mass has been gained — meaning energy must be supplied.

3. Plug in the given masses

We have:

$m(^{56}_{26}\text{Fe}) = 55.93494\ \text{u}$
$m(^{28}_{13}\text{Al}) = 27.98191\ \text{u}$

So:

Q = [55.93494 - (27.98191 + 27.98191)] \ \text{u} \times c^2

Q = [55.93494 - 55.96382] \ \text{u} \times c^2

Q = (-0.02888\ \text{u}) \times c^2

4. Convert to energy units

We know $1\ \text{u} = 931.5\ \text{MeV}/c^2$ . So:

Q = -0.02888 \times 931.5\ \text{MeV}

$Q \approx -26.9\ \text{MeV}$ …

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Common Mistakes in Nuclear Reaction Balancing (Fe Fission Problem)

Mistake 1: Forgetting to check charge conservation

Students often jump straight to the mass defect calculation without verifying that the reaction is even balanced. Here, the parent nucleus is $^{56}_{26}\text{Fe}$ with atomic number 26. Two $^{28}_{13}\text{Al}$ fragments give a total atomic number of $13 + 13 = 26$ — that checks out. But the mass numbers: $56$ vs $28 + 28 = 56$ — also fine. So the reaction is balanced.

How to avoid: Always write the full nuclear equation first, with both mass numbers and atomic numbers on each side. Tick them off one by one before doing any energy calculation.

Mistake 2: Using the wrong Q-value formula

The Q-value for fission is:

Q = \left[m_{\text{parent}} - \sum m_{\text{fragments}}\right]c^2

Some students write it as $Q = \sum m_{\text{products}} - m_{\text{reactant}}$ , which flips the sign. If you do that, you'll get a negative Q and wrongly conclude the reaction is impossible — or vice versa.

How to avoid: Remember: $Q = (\text{initial mass} - \text{final mass}) \times c^2$ . A positive Q means energy is released (exothermic/energetically possible). A negative Q means energy must be supplied.

Mistake 3: Forgetting to multiply fragment masses by the number of fragments

The problem says "two equal fragments." Each fragment is $^{28}_{13}\text{Al}$ with mass $27.98191\ \text{u}$ . The total final mass is $2 \times 27.98191\ \text{u}$ , not $27.98191\ \text{u}$ .

How to avoid: Circle the stoichiometric coefficients in your nuclear equation. If the equation says $2\ \text{Al}$ , then the mass contribution is $2 \times m(\text{Al})$ .

Mistake 4: Not converting atomic mass units to energy correctly

Even if you get the mass defect right, you need $1\ \text{u} = 931.5\ \text{MeV}/c^2$ . Some students use $931$ or forget the $c^2$ entirely.

How to avoid: Write the conversion factor explicitly: $Q(\text{MeV}) = \Delta m(\text{u}) \times 931.5\ \text{MeV/u}$ .

Mistake 5: Misinterpreting the sign of Q

Let's do the actual calculation:

\Delta m = 55.93494 - 2(27.98191) = 55.93494 - 55.96382 = -0.02888\ \text{u}

$Q = (-0.02888) \times 931.5 = -26.9\ \text{MeV}$ …

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The Core Idea: Why QQQ Tells Us If a Reaction "Goes"

Step-by-Step Calculation

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The Core Idea: Why $Q$ Tells Us If a Reaction "Goes"

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