calculate the energy released in nuclear reactions using E = c2∆m

Mass Defect and Nuclear Binding Energy

Learning Objective

By the end of this lesson you will be able to calculate the energy released in nuclear reactions using the relation $$E = c^{2}\,\Delta m$$ where $\Delta m$ is the mass defect.

Key Concepts

• Atomic mass unit (u): $1\ \text{u}=1.660539\times10^{-27}\ \text{kg}$.
• Mass defect ($\Delta m$): The difference between the sum of the masses of the separate nucleons and the actual mass of the nucleus.
• Binding energy ($E_b$): The energy equivalent of the mass defect, $E_b = \Delta m\,c^{2}$.
• Binding energy per nucleon: $E_b/A$, useful for comparing the stability of different nuclei.

Why Mass Defect Occurs

When protons and neutrons combine to form a nucleus, part of their total rest mass is converted into the energy that holds the nucleons together. This loss of mass is the mass defect.

Calculating Mass Defect

1. Write the nuclear reaction and identify the reactants and products.
2. Obtain the atomic masses (in u) of all reactants and products from a reliable table.
3. Calculate the total mass of reactants ($M_{\text{react}}$) and products ($M_{\text{prod}}$).
4. Find the mass defect: $$\Delta m = M_{\text{react}} - M_{\text{prod}}$$ (positive $\Delta m$ indicates mass loss and energy release).
5. Convert $\Delta m$ to kilograms: $$\Delta m\ (\text{kg}) = \Delta m\ (\text{u}) \times 1.660539\times10^{-27}\ \text{kg/u}$$
6. Calculate the released energy: $$E = c^{2}\,\Delta m$$ where $c = 2.998\times10^{8}\ \text{m s}^{-1}$.
7. Express $E$ in mega‑electronvolts (MeV) using $1\ \text{MeV}=1.60218\times10^{-13}\ \text{J}$.

Sample Atomic Mass Table

Nucleus	Atomic mass (u)	Number of protons (Z)	Number of neutrons (N)
$^{1}\!H$ (proton)	1.007825	1	0
$^{1}\!n$ (neutron)	1.008665	0	1
$^{4}\!He$	4.002603	2	2
$^{235}\!U$	235.043930	92	143
$^{141}\!Ba$	140.914411	56	85
$^{92}\!Kr$	91.926156	36	56
$^{3}\!n$ (neutron)	1.008665	0	1

Worked Example 1 – Formation of Helium‑4

Reaction: $$2\,^{1}\!H + 2\,^{1}\!n \rightarrow \,^{4}\!He$$

1. Mass of reactants: $$M_{\text{react}} = 2(1.007825) + 2(1.008665) = 4.03298\ \text{u}$$
2. Mass of product: $$M_{\text{prod}} = 4.002603\ \text{u}$$
3. Mass defect: $$\Delta m = 4.03298 - 4.002603 = 0.030377\ \text{u}$$
4. Convert to kilograms: $$\Delta m = 0.030377 \times 1.660539\times10^{-27} = 5.04\times10^{-29}\ \text{kg}$$
5. Energy released: $$E = (2.998\times10^{8})^{2} \times 5.04\times10^{-29} = 4.52\times10^{-12}\ \text{J}$$
6. Convert to MeV: $$E = \frac{4.52\times10^{-12}}{1.60218\times10^{-13}} \approx 28.2\ \text{MeV}$$

The binding energy per nucleon for $^{4}\!He$ is $28.2\ \text{MeV}/4 \approx 7.05\ \text{MeV}$.

Worked Example 2 – Fission of $^{235}\!U$

Typical fission reaction:

$$^{235}\!U + ^{1}\!n \rightarrow \,^{141}\!Ba + \,^{92}\!Kr + 3\,^{1}\!n$$

1. Mass of reactants: $$M_{\text{react}} = 235.043930 + 1.008665 = 236.052595\ \text{u}$$
2. Mass of products: $$M_{\text{prod}} = 140.914411 + 91.926156 + 3(1.008665) = 235.966562\ \text{u}$$
3. Mass defect: $$\Delta m = 236.052595 - 235.966562 = 0.086033\ \text{u}$$
4. Convert to kilograms: $$\Delta m = 0.086033 \times 1.660539\times10^{-27} = 1.43\times10^{-28}\ \text{kg}$$
5. Energy released: $$E = (2.998\times10^{8})^{2} \times 1.43\times10^{-28} = 1.28\times10^{-11}\ \text{J}$$
6. Convert to MeV: $$E = \frac{1.28\times10^{-11}}{1.60218\times10^{-13}} \approx 200\ \text{MeV}$$

This energy appears as kinetic energy of the fragments, gamma radiation, and the kinetic energy of the emitted neutrons.

Summary Checklist

• Identify all reactants and products and write the balanced nuclear equation.
• Use a reliable atomic mass table (in atomic mass units).
• Calculate the mass defect $\Delta m$ in u, then convert to kg.
• Apply $E = c^{2}\Delta m$ and convert the result to MeV.
• Interpret the binding energy per nucleon to assess nuclear stability.

Common Pitfalls

• Forgetting to include the mass of the incident neutron in fission calculations.
• Using the mass of the neutral atom instead of the nuclear mass; the electron masses cancel when both sides contain the same number of electrons, but be consistent.
• Neglecting to convert the mass defect from atomic mass units to kilograms before applying $E = c^{2}\Delta m$.

Practice Questions

1. Calculate the energy released when two deuterium nuclei ($^{2}\!H$) fuse to form a helium‑3 nucleus ($^{3}\!He$) and a neutron.
2. Determine the binding energy per nucleon for $^{56}\!Fe$ given its atomic mass is $55.934937\ \text{u}$.
3. A sample of $^{239}\!Pu$ undergoes alpha decay: $$^{239}\!Pu \rightarrow \,^{235}\!U + \,^{4}\!He$$. Compute the energy released.

Further Reading

Consult the Cambridge International AS & A Level Physics syllabus (9702) for additional examples and the recommended tables of nuclear masses.