Q.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 $m\left(^{28}_{13}\text{Al}\right) = 27.98191\ \text{u}$.

Concept: Nuclear Reaction Balancing — The $Q$-value determines if a reaction is energetically possible: $Q > 0$ means energy is released (possible), $Q < 0$ means energy must be supplied (not possible).

Step 1: Write the reaction

Step 2: Compute the mass defect

Total mass of products = $2 \times 27.98191\ \text{u} = 55.96382\ \text{u}$

Mass of reactant = $55.93494\ \text{u}$

Mass defect $\Delta m = 55.93494 - 55.96382 = -0.02888\ \text{u}$

Step 3: Convert to $Q$-value …

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:

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.

Step-by-Step Calculation

1. Write the reaction clearly

We're considering:

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

2. Recall the $Q$-value formula

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

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:

4. Convert to energy units

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

$Q \approx -26.9\ \text{MeV}$ …

Method: Q-value calculation for nuclear reactions

The Q-value tells us whether a nuclear reaction is energetically possible. If $Q > 0$, the reaction is exothermic and can occur spontaneously. If $Q < 0$, energy must be supplied from outside.

Step 1: Write the reaction clearly

The proposed fission is:

Step 2: Write the Q-value formula

For a reaction $\text{Parent} \rightarrow \text{Daughter}_1 + \text{Daughter}_2$, the Q-value is:

In atomic mass units, we can work directly in u, since $1\ \text{u} \cdot c^2 \approx 931.5\ \text{MeV}$.

Step 3: Plug in the given masses

So:

Step 4: Compute the mass difference

Step 5: Convert to energy

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

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:

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:

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