Mass-Energy and Nuclear Binding Energy

Q: What is mass defect in nuclear physics?

Mass defect is the difference between the total mass of the separate protons and neutrons in a nucleus and the actual mass of the nucleus. The nucleus is always lighter. For helium-4 the mass defect is 0.0304 u, which corresponds to its binding energy of 28.3 MeV.

Q: What is binding energy of a nucleus?

Binding energy is the minimum energy needed to break a nucleus into its free protons and neutrons, or equally the energy released when those nucleons come together. It equals the mass defect times c², and one atomic mass unit of mass defect gives 931.5 MeV.

Q: Why is binding energy per nucleon a measure of nuclear stability?

Total binding energy grows with the number of nucleons, so it cannot compare nuclei of different sizes. Binding energy per nucleon is the average energy needed to remove one nucleon. The larger it is, the more tightly each nucleon is held and the more stable the nucleus.

Q: Which nucleus is the most stable?

Nuclei near iron and nickel have the highest binding energy per nucleon, about 8.8 MeV. Iron-56 is usually quoted, while nickel-62 is very slightly higher. Energy is released when lighter nuclei fuse or heavier nuclei split, because both move towards this peak.

Q: Why is nuclear density the same for all nuclei?

The mass of a nucleus is proportional to its mass number, and its radius follows R = R₀A^1/3, so its volume is also proportional to the mass number. The two cancel and every nucleus has a density of about 2.3 times ten to the seventeen kilogram per cubic metre.

Q: What is the value of 1 atomic mass unit in MeV?

One atomic mass unit is one twelfth of the mass of a carbon-12 atom, 1.6605 times ten to the minus 27 kilogram. Its energy equivalent from E = mc² is 931.494 MeV, usually rounded to 931.5 MeV in calculations.

Q: How is binding energy asked in NEET?

NEET questions ask for the binding energy or mass defect from given masses, compare stability using binding energy per nucleon, use the radius formula R = R₀A^1/3 and nuclear density, and read the shape of the binding energy per nucleon graph. Remember 931.5 MeV per u.

Q: How is binding energy tested in JEE Main and Advanced?

JEE Main asks mass defect, binding energy per nucleon and energy released from the curve. JEE Advanced adds Q values with atomic masses, separation energies, beta plus mass corrections and questions based on the liquid drop formula and its surface and Coulomb terms.

PhysicsNucleiFor NEET aspirants

Mass-energy and nuclear binding energy explain why a nucleus holds together and where nuclear energy comes from. A nucleus always weighs less than its separate protons and neutrons. This missing mass, the mass defect $Δ m$ , is stored as binding energy $Δ m c^{2}$ , with $1 u = 931.5 MeV$ . Binding energy per nucleon peaks near $A = 56$ at about $8.8 MeV$ , so both fission and fusion release energy. Mass-energy and nuclear binding energy are tested every year in JEE Main and NEET.

On this page1The nucleus2Isotopes, isobars, isotones3Atomic mass unit4Size and density5Mass-energy equivalence6Mass defect7Binding energy8BE per nucleon curve9Q value10Liquid drop model

Key Formulas - Quick Reference

Notation $_{Z}^{A} X$ : $Z$ protons, $N = A - Z$ neutrons, $A$ nucleons
★ Must learn $1 u = \frac{1}{12}$ mass of a $^{12} C$ atom $= 1.6605 \times 1 0^{- 27} kg = 931.5 MeV / c^{2}$
★ Must learnNuclear radius $R = R_{0} A^{1/3}$ , $R_{0} \approx 1.2 fm$ $(1 fm = 1 0^{- 15} m)$ ; density $ρ = \frac{3 m _{u}}{4 π R _{0}^{3}} \approx 2.3 \times 1 0^{17} kg m^{- 3}$ , the same for all nuclei
Mass-energy equivalence: $E = m c^{2}$ ; $1 kg \equiv 9 \times 1 0^{16} J$
★ Must learnMass defect $Δ m = Z m_{p} + N m_{n} - M_{nucleus} = Z m_{H} + N m_{n} - M_{atom}$
★ Must learnBinding energy $B E = Δ m c^{2} = Δ m (in u) \times 931.5 MeV$ ; stability grows with $\frac{B E}{A}$
Separation energies: $S_{n} = B E (^{A} X) - B E (^{A - 1} X)$ , $S_{p} = B E (_{Z}^{A} X) - B E (_{Z - 1}^{A - 1} Y)$
★ Must learnQ value: $Q = (\sum m_{initial} - \sum m_{final}) c^{2} = \sum B E_{final} - \sum B E_{initial}$

1. The Nucleus and Its Constituents

Rutherford's $α$ -scattering experiment (1911) showed that the positive charge and almost all the mass of an atom sit in a tiny central nucleus. The nucleus is made of two kinds of particles, together called nucleons:

Particle	Charge	Mass (u)	Mass (kg)	Discovered
Proton $p$	$+ e$	$1.007276$	$1.6726 \times 1 0^{- 27}$	Rutherford (1919), as the hydrogen nucleus
Neutron $n$	$0$	$1.008665$	$1.6749 \times 1 0^{- 27}$	James Chadwick (1932)
Electron $e$	$- e$	$0.000549$	$9.109 \times 1 0^{- 31}$	J. J. Thomson (1897)

The neutron is slightly heavier than the proton ( $m_{n} - m_{p} \approx 1.29 MeV / c^{2}$ ), which is why a free neutron can decay into a proton. A nucleus is labelled by two whole numbers:

Atomic number $Z$ : the number of protons. It fixes the chemical element.
Mass number $A$ : the total number of nucleons, $A = Z + N$ , where $N$ is the neutron number. $A$ is the whole number nearest to the atomic mass in u.

Figure 1: A nucleus is about

1 0^{5}

times smaller than its atom but carries almost all its mass. The symbol

_{Z}^{A} X

gives the mass number

A

(nucleons) and atomic number

Z

(protons); the neutron number is

N = A - Z

The nucleus is written $_{Z}^{A} X$ (some books write $_{Z} X^{A}$ ). For example $_{79}^{197} Au$ has $79$ protons and $197 - 79 = 118$ neutrons. A particular nucleus with given $Z$ and $N$ is called a nuclide.

1.1 Isotopes, isobars and isotones

Name	Same	Different	Examples
Isotopes	$Z$ (same element)	$N$ and $A$	$_{1}^{1} H$ , $_{1}^{2} H$ , $_{1}^{3} H$ ; $_{17}^{35} Cl$ , $_{17}^{37} Cl$
Isobars	$A$	$Z$ and $N$	$_{1}^{3} H$ , $_{2}^{3} He$ ; $_{6}^{14} C$ , $_{7}^{14} N$
Isotones	$N$	$Z$ and $A$	$_{1}^{3} H$ , $_{2}^{4} He$ ( $N = 2$ ); $_{6}^{14} C$ , $_{8}^{16} O$ ( $N = 8$ )
Isodiaphers	$N - Z$	$Z$ , $A$	$_{92}^{238} U$ and $_{90}^{234} Th$ ( $N - Z = 54$ ): parent and daughter in $α$ -decay

Figure 2: A corner of the chart of nuclides. Isotopes lie in a column (same

Z

), isotones in a row (same

N

) and isobars on a diagonal (same

A = Z + N

). Only the peach cells are stable.

Exam Trick

Match the letter to what stays the same. Isotopes: same protons. Isotones: same neutrons. Isobars: same mass number $A$ . Isotopes are chemically identical because chemistry depends only on $Z$ .

2. Atomic Mass Unit and Atomic Masses

Nuclear masses are tiny in kilograms, so they are measured in the atomic mass unit (u), defined from carbon-12:

1 u = \frac{mass of one ^{12} C atom}{12} = \frac{12 \times 1 0 ^{- 3} kg mol ^{- 1}}{12 \times N _{A}} = 1.660539 \times 1 0^{- 27} kg

Its energy equivalent is $1 u \times c^{2} = 1.4924 \times 1 0^{- 10} J = 931.494 MeV$ . Exam problems use $931.5 MeV$ per u (sometimes $931 MeV$ ).

Tables list atomic masses (nucleus plus $Z$ electrons), because these are what mass spectrometers measure. On this scale $m_{p} = 1.007276 u$ , $m_{n} = 1.008665 u$ , and the hydrogen atom $m_{H} = 1.007825 u$ . The atomic mass of an element in the periodic table (for example $35.45 u$ for chlorine) is an average over its isotopes weighted by their abundance: $0.758 \times 34.97 + 0.242 \times 36.97 = 35.45 u$ .

Key idea

The nucleus is labelled by

Z

and

A

; masses are in u, and every u of mass is worth

931.5 MeV

of energy.

3. Size and Density of the Nucleus

Scattering of fast electrons and $α$ -particles shows that nuclei are roughly spherical, with a radius that grows as the cube root of the mass number:

R = R_{0} A^{1/3}, R_{0} \approx 1.2 fm = 1.2 \times 1 0^{- 15} m

The femtometre (fermi), $1 fm = 1 0^{- 15} m$ , is the natural unit. Different experiments give $R_{0}$ between $1.1$ and $1.3 fm$ ; NCERT uses $1.2 fm$ . So volume $\propto R^{3} \propto A$ .

Figure 3:

R = R_{0} A^{1/3}

with

R_{0} = 1.2 fm

(orange) and

1.1 fm

(dashed). Going from

^{4} He

^{238} U

multiplies

A

by about

60

but

R

only by

6 0^{1/3} \approx 3.9

(1.90 to 7.44 fm).

3.1 Nuclear density is the same for all nuclei

Mass of a nucleus $\approx A m_{u}$ (each nucleon weighs about $1 u$ ).
Volume $= \frac{4}{3} π R^{3} = \frac{4}{3} π R_{0}^{3} A$ .
Density:
$ρ = \frac{A m _{u}}{\frac{4}{3} π R _{0}^{3} A} = \frac{3 m _{u}}{4 π R _{0}^{3}} = \frac{3 \times 1.66 \times 1 0 ^{- 27}}{4 π \times ( 1.2 \times 1 0 ^{- 15} ) ^{3}} \approx 2.3 \times 1 0^{17} kg m^{- 3}$

Figure 4: Nuclei drawn to scale (

R_{0} = 1.2 fm

). The volume

\frac{4}{3} π R_{0}^{3} A

grows exactly as the mass

A m_{u}

, so all nuclei have the same density

ρ = \frac{3 m _{u}}{4 π R _{0}^{3}} = 2.29 \times 1 0^{17} kg m^{- 3}

$A$ cancels, so all nuclei, from helium to uranium, have the same density. This is about $1 0^{14}$ times the density of water: a teaspoon of nuclear matter would weigh some billion tonnes. Neutron stars are made of matter at this density. A constant density is what we expect of a liquid drop: nucleons are packed like molecules in a drop, which is the idea behind the liquid drop model (Section 9).

Quick Recall: tap to check

By what factor does the radius grow from

^{27} Al

^{216} Po

(\frac{216}{27})^{1/3} = 8^{1/3} = 2

Does nuclear density depend on the mass number?

No. Mass

\propto A

and volume

\propto A

, so

ρ = \frac{3 m _{u}}{4 π R _{0}^{3}}

is the same for every nucleus.

What is

1 fm

1 0^{- 15} m

, the natural unit of nuclear size.

4. Mass-Energy Equivalence

Einstein's special relativity (1905) showed that mass is a form of energy. A body of mass $m$ at rest has rest-mass energy:

E = m c^{2}

In any process the total energy, rest-mass energy included, is conserved. When energy is released, the total rest mass of the system falls by $Δ m = \frac{Δ E}{c ^{2}}$ .

One kilogram is equivalent to $1 \times (3 \times 1 0^{8})^{2} = 9 \times 1 0^{16} J$ . In chemical reactions the mass change is only about $1 0^{- 9}$ of the mass and cannot be measured, so chemistry uses "conservation of mass". In nuclear reactions about $0.1%$ of the mass changes into energy, and the change is easily seen. Pair annihilation ( $e^{-} + e^{+} \to 2 γ$ , each photon $0.511 MeV$ ) turns rest mass completely into radiation, and pair production does the reverse.

Figure 5:

E = m c^{2}

with nothing hidden. The electron's rest energy is

m_{e} c^{2} = 0.00054858 \times 931.494 = 0.511 MeV

. Annihilation turns the whole rest mass of

e^{-}

and

e^{+}

into two

γ

-rays; pair production needs a photon of at least

2 m_{e} c^{2} = 1.022 MeV

Exam Trick

Never convert to joules unless asked. Keep masses in u and multiply by $931.5$ to get MeV directly. Then $1 MeV = 1.6 \times 1 0^{- 13} J$ if a joule answer is needed.

5. Mass Defect

Adding up the masses of the protons and neutrons in a nucleus always gives more than the measured mass of the nucleus. The difference is the mass defect.

Δ m = [Z m_{p} + (A - Z) m_{n}] - M_{nucleus}

Since $M_{nucleus} = M_{atom} - Z m_{e}$ (neglecting the tiny electron binding energies), and $m_{p} + m_{e} = m_{H}$ , the same result in terms of atomic masses is

Δ m = [Z m_{H} + (A - Z) m_{n}] - M_{atom}

Use $m_{H}$ with atomic masses and the electron masses cancel automatically.

Figure 6: The mass defect of

^{4} He

2 m_{p} + 2 m_{n} = 4.031883 u

but the nucleus has

m = 4.001506 u

. The missing

Δ m = 0.03038 u

is the binding energy

Δ m c^{2} = 28.3 MeV

(the axis is broken to show the small difference).

For helium-4: $2 m_{H} + 2 m_{n} = 2 (1.007825) + 2 (1.008665) = 4.032980 u$ , while $M (^{4} He) = 4.002603 u$ , so $Δ m = 0.030377 u$ . The nucleus is lighter because energy was given out when it formed: the mass defect is the mass of that energy.

6. Binding Energy

The binding energy of a nucleus is the minimum energy needed to separate it completely into its free protons and neutrons (at rest, far apart). Equally, it is the energy released when the free nucleons combine to form the nucleus.

B E = Δ m c^{2} = Δ m (in u) \times 931.5 MeV

For $^{4} He$ : $B E = 0.030377 \times 931.5 = 28.3 MeV$ .

Figure 7: Binding energy two ways. It is the energy needed to pull a nucleus apart into free nucleons, and also the energy given out when those nucleons come together. Both equal

Δ m c^{2}

The binding energy is the depth of the energy well in which the nucleons sit. The larger it is, the harder the nucleus is to break up. A few useful values:

Nucleus	$Δ m$ (u)	$B E$ (MeV)	$\frac{B E}{A}$ (MeV)
$_{1}^{2} H$ (deuteron)	0.002388	2.22	1.11
$_{2}^{4} He$	0.030377	28.30	7.07
$_{6}^{12} C$	0.098940	92.16	7.68
$_{8}^{16} O$	0.137005	127.62	7.98
$_{26}^{56} Fe$	0.528463	492.3	8.79
$_{92}^{235} U$	1.915058	1783.9	7.59
$_{92}^{238} U$	1.934194	1801.7	7.57

6.1 Separation energy

The energy needed to remove just one nucleon is the separation energy. Removing a neutron from $_{Z}^{A} X$ leaves $_{Z}^{A - 1} X$ , so

S_{n} = [m (_{Z}^{A - 1} X) + m_{n} - m (_{Z}^{A} X)] c^{2} = B E (_{Z}^{A} X) - B E (_{Z}^{A - 1} X)

and for a proton $S_{p} = [m (_{Z - 1}^{A - 1} Y) + m_{H} - m (_{Z}^{A} X)] c^{2}$ . For $^{17} O$ , $S_{n} = 4.14 MeV$ , much less than the $7.75 MeV$ average, because the odd neutron is loosely bound outside the tightly bound $^{16} O$ core.

Binding energy (BE)

Energy to pull the whole nucleus apart into $A$ free nucleons. Grows with size: $28.3 MeV$ for He-4, $1802 MeV$ for U-238.

Binding energy per nucleon (BE/A)

Average energy to remove one nucleon. It measures stability: Fe-56 ( $8.79 MeV$ ) is more stable than U-238 ( $7.57 MeV$ ) even though its total BE is far smaller.

Quick Recall: tap to check

Why is a nucleus lighter than its free nucleons?

Energy

B E

was released when it formed; the lost mass is

\frac{B E}{c ^{2}}

Why use

m_{H}

rather than

m_{p}

with atomic masses?

m_{H} = m_{p} + m_{e}

includes one electron per proton, so the

Z

electrons in

M_{atom}

cancel.

Which is the better measure of stability: BE or BE/A?

\frac{B E}{A}

. A bigger nucleus has a bigger total BE simply because it has more nucleons.

7. Binding Energy per Nucleon Curve

Plotting $\frac{B E}{A}$ against $A$ for all stable nuclei gives one of the most important graphs in physics.

Figure 8: Binding energy per nucleon from measured masses (grey dots: all 288 naturally occurring nuclides; orange: the most abundant isotope of each element). The curve peaks at

^{56} Fe

(8.79 MeV), so joining light nuclei (fusion) or splitting heavy ones (fission) both release energy.

7.1 Features of the curve

Light nuclei ( $A < 20$ ): $\frac{B E}{A}$ is small and rises steeply, with sharp peaks for $^{4} He$ , $^{12} C$ , $^{16} O$ , $^{20} Ne$ (nuclei built of " $α$ -particle units").
Middle nuclei ( $30 < A < 170$ ): the curve is nearly flat at $8.0$ to $8.8 MeV$ . The maximum is about $8.8 MeV$ near $A = 56$ to $62$ ( $^{56} Fe$ : $8.79 MeV$ ; $^{62} Ni$ : $8.795 MeV$ is the true maximum). These are the most stable nuclei.
Heavy nuclei ( $A > 170$ ): $\frac{B E}{A}$ falls slowly, to $7.6 MeV$ for uranium, because Coulomb repulsion between many protons grows.

Figure 9: Light nuclei are irregular.

^{4} He

(7.07 MeV per nucleon) is far more tightly bound than its neighbours

^{3} He

(2.57) and

^{6} Li

(5.33); that is why heavy nuclei emit

α

-particles rather than single nucleons.

7.2 What the curve tells us

Fission: a heavy nucleus ( $\frac{B E}{A} \approx 7.6 MeV$ ) that splits into two middle nuclei ( $\approx 8.5 MeV$ ) gains about $0.9 MeV$ per nucleon: roughly $0.9 \times 236 \approx 200 MeV$ per fission.
Fusion: light nuclei joining into a heavier one climb the steep left side, releasing even more energy per nucleon (D-T fusion: $3.5 MeV$ per nucleon).
Short-range force: the flat middle shows that each nucleon feels only its near neighbours. If every nucleon attracted every other, $B E$ would grow as $A^{2}$ and $\frac{B E}{A}$ as $A$ . This saturation is taken up in the next concept, Nuclear Forces and Nuclear Energy.

Figure 10: Total binding energy grows almost in proportion to

A

(about

8 MeV

per nucleon), not as the number of pairs

\frac{A ( A - 1 )}{2}

. Each nucleon attracts only its near neighbours: the nuclear force is short-range and saturates.

Exam Trick

Energy released = (BE after) − (BE before). When binding energies per nucleon are given, compute $\sum A \cdot \frac{B E}{A}$ for products minus the same for reactants. Example: a nucleus with $A = 240$ , $\frac{B E}{A} = 7.6 MeV$ splits into two with $A = 120$ , $\frac{B E}{A} = 8.5 MeV$ : $Q = 240 (8.5 - 7.6) = 216 MeV$ .

Key idea

The higher the binding energy per nucleon, the more stable the nucleus. Energy is released whenever nucleons move to nuclei with higher

\frac{B E}{A}

: towards

A \approx 56

from either side.

8. Q Value of a Nuclear Reaction

For any nuclear reaction or decay $a + X \to Y + b$ , the Q value is the energy released:

Q = [(m_{a} + m_{X}) - (m_{Y} + m_{b})] c^{2} = (K_{Y} + K_{b}) - (K_{a} + K_{X})

Because the number of protons and neutrons does not change, $Q$ also equals (total BE of products) − (total BE of reactants). $Q > 0$ : exothermic, rest mass turns into kinetic energy. $Q < 0$ : endothermic, needs energy input.

Figure 11:

Q = (\sum m_{initial} - \sum m_{final}) c^{2}

(levels not to scale). D-T fusion:

Q = + 17.6 MeV

, rest mass turns into kinetic energy.

^{14} N (α, p)^{17} O

Q = - 1.19 MeV

, kinetic energy turns into rest mass, and momentum conservation raises the threshold to

∣ Q ∣ (1 + \frac{m _{α}}{m _{N}}) = 1.53 MeV

Charge and nucleon number balance, so atomic masses can be used on both sides and the electrons cancel. The one exception is $β^{+}$ decay, where $2 m_{e}$ must be subtracted (see Radioactivity). Example: $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He + n$ gives $Q = (2.014102 + 3.016049 - 4.002603 - 1.008665) \times 931.5 = 17.6 MeV$ .

JEE Advanced

Threshold energy of an endothermic reaction. For $a + X \to Y + b$ with $Q < 0$ and $X$ at rest, supplying just $∣ Q ∣$ is not enough: the centre of mass must keep moving, so part of the projectile's kinetic energy stays as kinetic energy. The minimum (threshold) kinetic energy of the projectile is

K_{th} = ∣ Q ∣ (1 + \frac{m _{a}}{m _{X}})

For Rutherford's reaction $^{14} N (α, p)^{17} O$ , $Q = 1.19 MeV$ but $K_{th} = 1.19 (1 + \frac{4}{14}) = 1.53 MeV$ (Figure 11).

Quick Recall: tap to check

What does a positive Q value mean?

Energy is released: the products have less rest mass and more kinetic energy than the reactants.

Can Q be found from binding energies alone?

Yes:

Q = \sum B E_{products} - \sum B E_{reactants}

, since the nucleons are the same on both sides.

In which decay must you subtract

2 m_{e}

when using atomic masses?

β^{+}

decay.

9. Liquid Drop Model and Semi-empirical Mass Formula

The constant density and the saturation of binding energy suggest treating the nucleus like a drop of incompressible liquid. Each effect on the binding energy then has a simple origin.

JEE Advanced

Bethe-Weizsäcker formula.

B E = a_{v} A - a_{s} A^{2/3} - a_{c} \frac{Z ( Z - 1 )}{A ^{1/3}} - a_{a} \frac{( N - Z ) ^{2}}{A} \pm δ

Volume term $a_{v} A$ ( $a_{v} \approx 15.8 MeV$ ): each nucleon bonds with its neighbours, so $B E \propto A$ .
Surface term $- a_{s} A^{2/3}$ ( $a_{s} \approx 17.8 MeV$ ): surface nucleons have fewer neighbours; surface area $\propto R^{2} \propto A^{2/3}$ . Most important for light nuclei.
Coulomb term $- a_{c} \frac{Z ( Z - 1 )}{A ^{1/3}}$ ( $a_{c} \approx 0.71 MeV$ ): proton-proton repulsion; energy of a charged sphere $\propto \frac{Q ^{2}}{R}$ . Most important for heavy nuclei.
Asymmetry term $- a_{a} \frac{( N - Z ) ^{2}}{A}$ ( $a_{a} \approx 23.7 MeV$ ): a quantum effect favouring $N = Z$ .
Pairing term $δ$ : even-even nuclei are extra bound, odd-odd less.

With these five terms the formula reproduces measured binding energies of heavy nuclei to within about $1%$ . For $^{56} Fe$ it gives 8.85 $MeV$ per nucleon against the measured 8.79 $MeV$ .

Figure 12: The semi-empirical (liquid drop) formula term by term, per nucleon, with

Z

set to its most stable value. Surface loss hurts light nuclei, Coulomb repulsion hurts heavy ones; the peak between them is near

A \approx 60

. Black dots are measured values.

Key idea

Surface loss pulls

\frac{B E}{A}

down for light nuclei, Coulomb repulsion pulls it down for heavy nuclei; the peak between them is near iron.

10. Problem-Solving Map and Revision

Use the flowchart to pick the correct masses, then the mind map to revise the concept.

Figure 13: Choosing the right masses. Using

m_{H}

with atomic masses makes the electrons cancel automatically.

Figure 14: Mind map of this concept. Read the left column, then the right; cover a branch, recall its three points, then check.

11. Solved Examples

Solved Example 1

Find the density of nuclear matter, taking

R_{0} = 1.1 \times 1 0^{- 15} m

and

m_{p} = 1.67 \times 1 0^{- 27} kg

. Repeat with

R_{0} = 1.2 fm

Solution:

$ρ = \frac{A m _{p}}{\frac{4}{3} π R _{0}^{3} A} = \frac{3 m _{p}}{4 π R _{0}^{3}}$ , independent of $A$ .

ρ = \frac{3 \times 1.67 \times 1 0 ^{- 27}}{4 \times 3.14 \times ( 1.1 \times 1 0 ^{- 15} ) ^{3}} = 3.0 \times 1 0^{17} kg m^{- 3}

With $R_{0} = 1.2 fm$ the cube is $1.728 \times 1 0^{- 45}$ and $ρ = 2.3 \times 1 0^{17} kg m^{- 3}$ .

Answer: $ρ \approx 3 \times 1 0^{17} kg m^{- 3}$ ( $R_{0} = 1.1 fm$ ), $2.3 \times 1 0^{17} kg m^{- 3}$ ( $R_{0} = 1.2 fm$ ), the same for every nucleus.

Solved Example 2

The radius of a germanium nucleus is measured to be twice the radius of

_{4}^{9} Be

. How many nucleons are there in the Ge nucleus?

Solution:

$R \propto A^{1/3}$ , so $\frac{A _{G e}}{A _{B e}} = (\frac{R _{G e}}{R _{B e}})^{3} = 2^{3} = 8$ .

Answer: $A_{G e} = 8 \times 9 = 72$ nucleons ( $^{72} Ge$ ).

Solved Example 3

Calculate the energy equivalent of

1 g

of matter. For how long could it run a

1 kW

heater?

Solution:

$E = m c^{2} = 1 0^{- 3} \times (3 \times 1 0^{8})^{2} = 9 \times 1 0^{13} J$ .

Time $= \frac{9 \times 1 0 ^{13}}{1 0 ^{3}} = 9 \times 1 0^{10} s \approx 2850$ years.

Answer: $9 \times 1 0^{13} J$ ( $2.5 \times 1 0^{7} kWh$ ), enough for about 2850 years.

Solved Example 4

Find the binding energy and binding energy per nucleon of

_{8}^{16} O

. Atomic mass of

^{16} O = 15.994915 u

m_{H} = 1.007825 u

m_{n} = 1.008665 u

Solution:

$Δ m = 8 m_{H} + 8 m_{n} - M = 8 (1.007825) + 8 (1.008665) - 15.994915$

$= 8.062600 + 8.069320 - 15.994915 = 0.137005 u$ .

$B E = 0.137005 \times 931.5 = 127.6 MeV$ ; $\frac{B E}{A} = \frac{127.6}{16}$ .

Answer: $B E = 127.6 MeV$ , $\frac{B E}{A} = 7.98 MeV$ per nucleon.

Solved Example 5

Calculate the binding energy per nucleon of

_{26}^{56} Fe

. Atomic mass

= 55.934936 u

Solution:

$N = 30$ . $Δ m = 26 (1.007825) + 30 (1.008665) - 55.934936$

$= 26.203450 + 30.259950 - 55.934936 = 0.528464 u$ .

$B E = 0.528464 \times 931.5 = 492.3 MeV$ .

Answer: $\frac{B E}{A} = \frac{492.3}{56} = 8.79 MeV$ per nucleon, close to the maximum of the curve.

Solved Example 6

Find the energy needed to remove one neutron from

_{8}^{17} O

. Atomic masses:

^{17} O = 16.999132 u

^{16} O = 15.994915 u

m_{n} = 1.008665 u

Solution:

$^{17} O \to^{16} O + n$ . $S_{n} = [m (^{16} O) + m_{n} - m (^{17} O)] \times 931.5$

$= (15.994915 + 1.008665 - 16.999132) \times 931.5 = 0.004448 \times 931.5$ .

Answer: $S_{n} \approx 4.14 MeV$ , about half the average binding energy per nucleon.

Solved Example 7

A nucleus with mass number

240

and

\frac{B E}{A} = 7.6 MeV

breaks into two fragments, each with

A = 120

and

\frac{B E}{A} = 8.5 MeV

. The energy released is
(A)

0.9 MeV

(B)

108 MeV

(C)

216 MeV

(D)

2040 MeV

Solution:

Answer: (C). $Q = 2 \times 120 \times 8.5 - 240 \times 7.6 = 2040 - 1824 = 216 MeV$ . Option (A) is the gain per nucleon; (D) is the total BE of the fragments.

Solved Example 8

Two deuterons (

\frac{B E}{A} = 1.1 MeV

) fuse to form

^{4} He

(

\frac{B E}{A} = 7.07 MeV

). The energy released is about
(A)

5.97 MeV

(B)

11.9 MeV

(C)

23.9 MeV

(D)

28.3 MeV

Solution:

Answer: (C). $Q = 4 \times 7.07 - 2 \times (2 \times 1.1) = 28.28 - 4.4 = 23.9 MeV$ . Each deuteron has $A = 2$ , so its total BE is $2.2 MeV$ .

Solved Example 9

Which of the following pairs are isotones?
(A)

_{6}^{14} C

_{7}^{14} N

(B)

_{6}^{14} C

_{8}^{16} O

(C)

_{6}^{12} C

_{6}^{14} C

(D)

_{1}^{3} H

_{2}^{3} He

Solution:

Answer: (B). $N = 14 - 6 = 8$ and $N = 16 - 8 = 8$ . (A) and (D) are isobars; (C) are isotopes.

Solved Example 10

An electron and a positron, both nearly at rest, annihilate into two photons. Find the energy and wavelength of each photon (

m_{e} c^{2} = 0.511 MeV

h c = 1240 eV nm

Solution:

Total momentum is zero, so the two photons fly apart in opposite directions with equal energies. Energy conservation: $2 E_{γ} = 2 m_{e} c^{2}$ .

$E_{γ} = 0.511 MeV$ ; $λ = \frac{h c}{E _{γ}} = \frac{1240 eV nm}{0.511 \times 1 0 ^{6} eV} = 2.43 \times 1 0^{- 3} nm$ .

Answer: $E_{γ} = 0.511 MeV$ each, $λ = 2.43 pm$ (a $γ$ -ray). One photon alone is impossible: it could not carry away zero momentum.

Solved Example 11

For

^{14} N +^{4} He \to^{17} O +^{1} H

, find

Q

and the threshold kinetic energy of the

α

-particle (nitrogen at rest). Atomic masses:

^{14} N = 14.003074

^{4} He = 4.002603

^{17} O = 16.999132

^{1} H = 1.007825 u

Solution:

$\sum m_{i} = 18.005677 u$ , $\sum m_{f} = 18.006957 u$ , so $Δ m = - 0.001280 u$ and $Q = - 0.001280 \times 931.5 = - 1.19 MeV$ .

$K_{th} = ∣ Q ∣ (1 + \frac{m _{α}}{m _{N}}) = 1.19 \times (1 + \frac{4.0026}{14.0031}) = 1.19 \times 1.286$ .

Answer: $Q = - 1.19 MeV$ (endothermic); $K_{th} \approx 1.53 MeV$ .

Solved Example 12

The binding energies per nucleon of

_{3}^{7} Li

and

_{2}^{4} He

are

5.60 MeV

and

7.07 MeV

. In the reaction

_{3}^{7} Li +_{1}^{1} H \to 2_{2}^{4} He

the energy released is
(A)

8.7 MeV

(B)

17.3 MeV

(C)

28.3 MeV

(D)

39.2 MeV

Solution:

Answer: (B). The proton has no binding energy. $Q = 2 \times 4 \times 7.07 - 7 \times 5.60 = 56.56 - 39.20 = 17.36 \approx 17.3 MeV$ . (A) is the energy per $α$ -particle, (C) is the binding energy of one $^{4} He$ and (D) that of $^{7} Li$ .

Practice Questions

How many protons and neutrons are in $_{92}^{235} U$ ?Answer: $92$ protons, $143$ neutrons
Find the ratio of the radii of $^{27} Al$ and $^{125} Te$ .Answer: $3 : 5$
Find the mass defect and binding energy of the deuteron. $m (^{2} H) = 2.014102 u$ .Answer: $0.002388 u$ ; $2.22 MeV$
Find the binding energy per nucleon of $^{12} C$ (atomic mass exactly $12 u$ ).Answer: $Δ m = 0.098940 u$ , $B E = 92.2 MeV$ , $7.68 MeV$ per nucleon
How much mass is converted into energy when $1 kWh$ is produced?Answer: $4 \times 1 0^{- 11} kg$
Two nuclei have mass numbers in the ratio $1 : 8$ . Find the ratio of their nuclear densities.Answer: $1 : 1$
Find $Q$ for $^{2} H +^{2} H \to^{3} He + n$ . Masses: $2.014102$ , $3.016029$ , $1.008665 u$ .Answer: $3.27 MeV$

Common Mistakes to Avoid

Watch out

Mixing atomic and nuclear masses: using $m_{p}$ with an atomic mass leaves $Z$ electron masses uncancelled. Use $m_{H}$ with atomic masses, $m_{p}$ with nuclear masses.
Taking the mass of a nucleon as exactly $1 u$ in binding-energy problems. The whole answer lives in the fourth decimal place; keep all given digits.
Judging stability by total $B E$ . Use $\frac{B E}{A}$ : $^{238} U$ has the larger $B E$ but $^{56} Fe$ is far more stable.
Writing $R \propto A$ or $R \propto A^{1/2}$ . It is $R \propto A^{1/3}$ , so volume, not radius, is proportional to $A$ .
Thinking heavier nuclei are denser. Nuclear density is the same for all nuclei.
Using $1 u = 931.5 MeV$ with the mass in kg or grams. Convert to u first, or use $E = m c^{2}$ in SI units.
Saying mass is 'destroyed' in a nuclear reaction. Rest mass turns into kinetic energy; total energy (and mass-energy) is conserved.
Forgetting that the maximum of the $\frac{B E}{A}$ curve is only about $8.8 MeV$ and occurs near $A = 56$ , not at uranium or helium.

Frequently Asked Questions

What is mass defect in nuclear physics?

Mass defect is the difference between the total mass of the separate protons and neutrons in a nucleus and the actual mass of the nucleus. The nucleus is always lighter. For helium-4 the mass defect is 0.0304 u, which corresponds to its binding energy of 28.3 MeV.

What is binding energy of a nucleus?

Binding energy is the minimum energy needed to break a nucleus into its free protons and neutrons, or equally the energy released when those nucleons come together. It equals the mass defect times $c^{2}$ , and one atomic mass unit of mass defect gives 931.5 MeV.

Why is binding energy per nucleon a measure of nuclear stability?

Total binding energy grows with the number of nucleons, so it cannot compare nuclei of different sizes. Binding energy per nucleon is the average energy needed to remove one nucleon. The larger it is, the more tightly each nucleon is held and the more stable the nucleus.

Which nucleus is the most stable?

Nuclei near iron and nickel have the highest binding energy per nucleon, about 8.8 MeV. Iron-56 is usually quoted, while nickel-62 is very slightly higher. Energy is released when lighter nuclei fuse or heavier nuclei split, because both move towards this peak.

Why is nuclear density the same for all nuclei?

The mass of a nucleus is proportional to its mass number, and its radius follows $R = R_{0} A^{1/3}$ , so its volume is also proportional to the mass number. The two cancel and every nucleus has a density of about 2.3 times ten to the seventeen kilogram per cubic metre.

What is the value of 1 atomic mass unit in MeV?

One atomic mass unit is one twelfth of the mass of a carbon-12 atom, 1.6605 times ten to the minus 27 kilogram. Its energy equivalent from $E = m c^{2}$ is 931.494 MeV, usually rounded to 931.5 MeV in calculations.

How is binding energy asked in NEET?

NEET questions ask for the binding energy or mass defect from given masses, compare stability using binding energy per nucleon, use the radius formula $R = R_{0} A^{1/3}$ and nuclear density, and read the shape of the binding energy per nucleon graph. Remember 931.5 MeV per u.

How is binding energy tested in JEE Main and Advanced?

JEE Main asks mass defect, binding energy per nucleon and energy released from the curve. JEE Advanced adds Q values with atomic masses, separation energies, beta plus mass corrections and questions based on the liquid drop formula and its surface and Coulomb terms.

Previous year questions on Mass-Energy and Nuclear Binding Energy

19 questions from past papers, each with a step-by-step solution.

Show all 19 questions

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