Mass Defect & Binding Energy

PHYNuclei — Structure & Binding Energy

A nucleus always weighs slightly less than the sum of its free protons and neutrons; that missing mass — the mass defect $\Delta m$ — has been converted into the binding energy that holds the nucleus together, via $E_b = \Delta m\,c^2$ . Dividing by the number of nucleons gives the binding energy per nucleon, the single best measure of nuclear stability, peaking near $8.8\,\text{MeV}$ for iron.

The same bookkeeping yields the $Q$ -value of a reaction (energy released or absorbed), so ISC examines this topic for both stability comparisons and fission/fusion energy calculations, with $1\,\text{u} = 931.5\,\text{MeV}$ as the workhorse conversion.

Mass defect of a nucleus

\Delta m = \left[Z\,m_p + (A-Z)\,m_n\right] - M_{nucleus}

Z

is the number of protons,

A-Z

the number of neutrons,

m_p

and

m_n

their masses,

M_{nucleus}

the actual nuclear (or atomic) mass.

\Delta m

is positive for a bound nucleus. For

^4\text{He}

, use

Z=2

protons and

2

neutrons.

Binding energy and binding energy per nucleon

E_b = \Delta m\,c^2 = \Delta m\,(\text{u})\times 931.5\ \text{MeV}, \qquad \dfrac{E_b}{A}

E_b

is the total binding energy,

\Delta m

in atomic mass units,

A

the mass number. Multiplying

\Delta m

in u by

931.5

gives

E_b

directly in MeV; dividing by

A

gives the stability measure (about

7.07\,\text{MeV}

for

^4\text{He}

Q-value of a nuclear reaction

Q = \left(\sum m_{reactants} - \sum m_{products}\right)\times 931.5\ \text{MeV}

the masses are in u;

Q>0

means energy is released (exothermic, e.g. fusion or fission),

Q<0

means energy is absorbed. For the D-T reaction

^2\text{H}+{}^3\text{H}\to{}^4\text{He}+n

, the mass lost reappears as kinetic energy of the products.

Mass-energy equivalence

E = mc^2, \qquad 1\,\text{u} = 931.5\ \text{MeV}/c^2

E

is the energy equivalent of mass

m

c

the speed of light. The conversion

1\,\text{u} \leftrightarrow 931.5\,\text{MeV}

replaces an explicit

c^2

whenever masses are given in u — the standard ISC route.

Always count nucleons correctly: $Z$ protons and $(A-Z)$ neutrons. For $^{16}\text{O}$ that is $8$ protons and $8$ neutrons; for $^4\text{He}$ , $2$ and $2$ .
When atomic masses (electrons included) are used consistently on both sides, the electron masses cancel, so you may use $m_p = 1.007825\,\text{u}$ (the hydrogen atom mass) without separate electron corrections.
Binding energy per nucleon, not total binding energy, ranks stability: $^4\text{He}$ has a large $E_b/A$ for a light nucleus, which is why the $\alpha$ -particle is exceptionally stable.
The $E_b/A$ curve rises sharply for light nuclei, peaks near $8.8\,\text{MeV}$ at iron ( $A\approx 56$ ), then falls slowly — fusion (light) and fission (heavy) both release energy by climbing toward the peak.
A positive $Q$ means the products are lighter than the reactants and the lost mass appears as kinetic energy; a negative $Q$ reaction needs an energy input to proceed.
Multiply the mass difference in u by $931.5$ to get MeV in one step — no need to handle $c^2$ in SI units explicitly.
Keep enough decimal places: mass defects are differences of nearly equal numbers (around $4.00$ u for helium), so rounding masses too early destroys the answer.

Where the marks go

Subtracting in the wrong order: $\Delta m = (\text{sum of nucleon masses}) - M_{nucleus}$ must be positive; reversing it gives a negative, meaningless mass defect.
Miscounting neutrons as $A$ instead of $A-Z$ , which throws off both the mass defect and the binding energy.
Rounding the input masses too soon — the defect is a tiny difference of large numbers, so premature rounding can swing $E_b/A$ by an MeV or more.
Confusing total binding energy with binding energy per nucleon when judging stability, or using $931.5\,\text{keV}$ instead of $931.5\,\text{MeV}$ for $1\,\text{u}$ .

How the board asks it

Numericalmass defect and binding energy via $E_b = \Delta m\,c^2$
The mass of a $^{16}_{8}\text{O}$ nucleus is $15.99491\,\text{u}$ . Given $m_p = 1.007825\,\text{u}$ and $m_n = 1.008665\,\text{u}$ , calculate the mass defect, the binding energy and the binding energy per nucleon. Take $1\,\text{u} = 931.5\,\text{MeV}$ .
Numericalq-value from mass difference
Calculate the energy released (in $\text{MeV}$ ) when four $^1_1\text{H}$ nuclei fuse to form a $^4_2\text{He}$ nucleus, given $m(^1_1\text{H}) = 1.007825\,\text{u}$ and $m(^4_2\text{He}) = 4.002603\,\text{u}$ and $1\,\text{u} = 931.5\,\text{MeV}$ .
Diagram / graphthe $E_b/A$ curve peaking near iron
Draw a graph showing the variation of binding energy per nucleon with mass number $A$ , mark the position of the most stable nucleus, and explain how the shape of the curve accounts for the energy released in nuclear fission and fusion.
Define / statedefinition of mass defect
Define the term 'mass defect' of a nucleus and write the relation connecting it to the binding energy of the nucleus.
Give reasons $E_b/A$ ranks stability
Two nuclei $X$ and $Y$ have binding energies per nucleon of $7.6\,\text{MeV}$ and $8.5\,\text{MeV}$ respectively. Giving a reason, state which nucleus is more stable.

Practise this topic

PreviousNuclear Size & Density NextMultiple Choice & Assertion-Reason

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.