Mass-Energy and Nuclear Binding Energy

Mass–energy equivalence

• Before special relativity, mass and energy were believed to be conserved separately. Einstein showed that mass must be treated as another form of energy, so that mass-energy can be converted into other forms of energy and back.

$$E = mc^2$$

• $E$ = energy equivalent of a mass $m$; $c = 3\times10^{8}$ m s$^{-1}$.
• Since mass and energy are interconvertible, we can no longer speak of separate conservation laws for mass and for energy — only the unified law of conservation of mass-energy holds. The most convincing evidence for this comes from nuclear physics.
• Worked (NCERT 13.2): energy equivalent of $1$ g of substance $= 10^{-3}\times(3\times10^{8})^2 = 9\times10^{13}$ J. A gram of matter, fully converted, releases an enormous amount of energy.

The energy equivalent of 1 u — a result to know cold

$$1\ \text{u} = 1.6605\times10^{-27}\ \text{kg}$$

$$E = 1.6605\times10^{-27}\times(2.9979\times10^{8})^2 = 1.4924\times10^{-10}\ \text{J}$$

$$E = \dfrac{1.4924\times10^{-10}}{1.602\times10^{-19}}\ \text{eV} = 0.9315\times10^{9}\ \text{eV} = 931.5\ \text{MeV}$$

$$\boxed{1\ \text{u} = 931.5\ \text{MeV}/c^2}$$

• Consequence: any mass defect written in u becomes an energy in MeV by multiplying by $931.5$. Do not also multiply by $c^2$ — the factor $931.5$ already contains it. (Some CBSE papers supply $931$ MeV instead of $931.5$; use whichever the paper gives.)

Mass defect

• One might expect the mass of a nucleus to equal the sum of the masses of its protons and neutrons. Experimentally the nuclear mass $M$ is ALWAYS less than that sum.
• NCERT's worked case, ${}^{16}_{8}\text{O}$ (8 protons, 8 neutrons):
• Mass of 8 neutrons $= 8\times1.00866$ u; mass of 8 protons $= 8\times1.00727$ u; mass of 8 electrons $= 8\times0.00055$ u; so the expected mass of the neutral atom is $8\times2.01593 = 16.12744$ u.
• Mass spectroscopy gives the atomic mass of ${}^{16}_{8}\text{O}$ as $15.99493$ u; subtracting the 8 electron masses gives the nuclear mass $15.99053$ u.
• The nucleus is therefore lighter than its constituents by $0.13691$ u. That shortfall is the MASS DEFECT.

$$\Delta M = \left[Z m_p + (A - Z)m_n\right] - M$$

• $Z m_p$ = mass of all the protons, $(A-Z)m_n$ = mass of all the neutrons, $M$ = actual nuclear mass.
• If the data are ATOMIC masses (which is how CBSE usually supplies them), use

$$\Delta M = \left[Z m_H + (A - Z)m_n\right] - M_{\text{atom}}$$

since the $Z$ electron masses cancel between the two sides.

• Why the mass is less — the 1-mark explanation: when protons and neutrons are brought together to within about $10^{-14}$ m, against their mutual Coulomb repulsion, to be bound by the strong nuclear force, energy is released. That energy is supplied at the expense of their rest mass, so the bound nucleus weighs less than the free nucleons. The deficit is the mass defect.

Binding energy

• By $E = mc^2$, the nucleus has less energy than its separated constituents by

$$E_b = \Delta M\,c^2$$

• Two equivalent statements of what $E_b$ means: (i) the energy RELEASED when free nucleons combine to form the nucleus; (ii) the energy that must be SUPPLIED to separate the nucleus completely into its free nucleons. Although we cannot literally tear a nucleus apart, $E_b$ is a convenient measure of how well it is held together.
• In practical units:

$$E_b(\text{MeV}) = \Delta M(\text{in u})\times931.5$$

• For ${}^{16}_{8}\text{O}$: $\Delta M = 0.13691$ u, so $E_b = 0.13691\times931.5 = 127.5$ MeV. That is the energy needed to break ${}^{16}_{8}\text{O}$ into its 8 protons and 8 neutrons.

Binding energy per nucleon

$$E_{bn} = \dfrac{E_b}{A}$$

• $E_{bn}$ is the AVERAGE energy per nucleon needed to separate the nucleus into individual nucleons. It, not $E_b$, is the measure of stability: the larger $E_{bn}$, the more tightly bound and the more stable the nucleus.
• MCQ shortcut: if the mass difference between ${}^{7}X$ and its constituent nucleons is $21.00$ u, then $E_b$ corresponds to $21.00$ u and $E_{bn}$ corresponds to $\dfrac{21.00}{7} = 3.0$ u — i.e. the energy equivalent of $3$ u.

The binding-energy-per-nucleon curve

The plot, in words

• Axes: $E_{bn}$ in MeV per nucleon (vertical, 0 to about 9) against mass number $A$ (horizontal, 0 to about 240).
• Shape: nearly zero at $A = 1$; a steep rise through the light nuclei, with sharp local peaks at ${}^{4}\text{He}$, ${}^{12}\text{C}$ and ${}^{16}\text{O}$; a broad, almost flat maximum of about $8.75$ MeV near $A = 56$ (iron); then a long, slow, smooth fall to $7.6$ MeV at $A = 238$ (uranium). Overall: a shallow, asymmetric hump.
• Mark $A = 30$ and $A = 170$ on the axis to show the flat middle region, and mark the peak at $A = 56$ and the value $7.6$ MeV at $A = 238$.

The two salient features CBSE asks for

• (i) $E_{bn}$ is practically constant — practically independent of the atomic/mass number — for nuclei of middle mass number, $30 < A < 170$. The curve has a maximum of about $8.75$ MeV at $A = 56$ and a value of $7.6$ MeV at $A = 238$.
• (ii) $E_{bn}$ is LOWER for both light nuclei ($A < 30$) and heavy nuclei ($A > 170$).

The four conclusions drawn from it

• (i) The nuclear force is attractive and strong enough to produce a binding energy of a few MeV per nucleon.
• (ii) The constancy of $E_{bn}$ over $30 < A < 170$ follows from the SHORT RANGE of the nuclear force. A nucleon deep inside a large nucleus feels only the neighbours that lie within the range of the force. If a nucleon has at most $p$ neighbours within range, its binding energy is proportional to $p$; write it as $pk$, with $k$ a constant of the dimensions of energy. Adding more nucleons does not change the binding of a nucleon already inside, and in a large nucleus most nucleons are inside rather than on the surface, so $E_{bn}$ stays constant at approximately $pk$. This property — that a nucleon influences only nucleons close to it — is the SATURATION property of the nuclear force.
• (iii) A very heavy nucleus, say $A = 240$, has a lower $E_{bn}$ than a nucleus with $A = 120$. So if the $A = 240$ nucleus breaks into two $A = 120$ nuclei, the nucleons become more tightly bound and energy is released. This is FISSION.
• (iv) Two very light nuclei ($A \le 10$) joining to form a heavier nucleus give a product with a higher $E_{bn}$ than the reactants; the final system is more tightly bound, so again energy is released. This is FUSION, and it is the energy source of the Sun.

Why the middle of the curve means "most stable"

• $E_{bn}$ measures directly how tightly each nucleon is held: the more energy needed per nucleon to pull the nucleus apart, the more stable it is.
• Nuclei on the flat middle portion ($30 < A < 170$) have the largest $E_{bn}$ of all, so they are the most tightly bound and the most stable. Everything else can lower its energy by moving TOWARDS this region — light nuclei by fusing up into it, heavy nuclei by fissioning down into it.
• The local peaks at ${}^{4}\text{He}$, ${}^{16}\text{O}$ and similar nuclides, on an otherwise smooth curve, are regarded as evidence of an atom-like shell structure inside nuclei.

Mass converted to energy while nucleon number is conserved

• The question CBSE repeats: if the number of protons and the number of neutrons are separately conserved in a nuclear reaction, in what sense is mass converted into energy?
• Answer to write: the binding energy of a nucleus makes a NEGATIVE contribution to its mass (that is the mass defect). Since proton number and neutron number are conserved, the total rest mass of the free nucleons is the same on both sides. But the total BINDING energy of the nuclei on the left need not equal that on the right. The difference in binding energies appears as the energy released or absorbed. Because binding energy contributes to mass, we say the difference in the total mass of the nuclei on the two sides has been converted into energy, or vice versa.
• Example to quote: ${}^{235}_{92}\text{U} + {}^{1}_{0}\text{n} \to {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\,{}^{1}_{0}\text{n} + Q$. Nucleon number is the same on both sides, yet the binding energies (hence the masses) differ, and the difference appears as $Q \approx 200$ MeV.
• The same logic applies to CHEMICAL reactions: the energy released there traces to a difference in chemical binding energies, which also contributes a (tiny) mass defect. Chemical mass defects are about a million times smaller than nuclear ones, which is why the false impression arises that mass-energy interconversion happens only in nuclear reactions.

Worked binding-energy numericals

Alpha particle (board, 3 marks)

• ${}^{4}_{2}\text{He}$ has 2 protons and 2 neutrons, so

$$\Delta m = 2m_p + 2m_n - m_{He}$$

• With standard NCERT data ($m_H = 1.007825$ u, $m_n = 1.008665$ u, atomic $m(^4\text{He}) = 4.002603$ u):

$$\Delta m = 2(1.007825) + 2(1.008665) - 4.002603 = 0.030377\ \text{u}$$

$$E_b = 0.030377\times931.5 \approx 28.3\ \text{MeV}, \qquad E_{bn} \approx 7.07\ \text{MeV}$$

• Note on the 2023 paper: it printed the proton mass as $1.0007828$ u and the helium mass as $4.002800$ u with $1\ \text{u} = 931$ MeV, giving $\Delta m = 0.016095$ u and $E_b \approx 14.98$ MeV. That is the answer the marking scheme wanted from the data as printed; the physically correct binding energy of an $\alpha$-particle is $28.3$ MeV. Always substitute the data the paper supplies, and show the method — the method is where the marks are.

Nitrogen (NCERT 13.1)

• ${}^{14}_{7}\text{N}$: $Z = 7$, $N = 7$, $m = 14.00307$ u.
• $\Delta m = 7(1.007825) + 7(1.008665) - 14.00307 = 0.11236$ u.
• $E_b = 0.11236\times931.5 = 104.7$ MeV; $E_{bn} = 7.48$ MeV.

Iron and bismuth (NCERT 13.2)

• ${}^{56}_{26}\text{Fe}$, $m = 55.934939$ u: $\Delta m = 26(1.007825) + 30(1.008665) - 55.934939 = 0.528461$ u, so $E_b = 492.3$ MeV and $E_{bn} = 8.79$ MeV.
• ${}^{209}_{83}\text{Bi}$, $m = 208.980388$ u: $\Delta m = 83(1.007825) + 126(1.008665) - 208.980388 = 1.760877$ u, so $E_b = 1640$ MeV and $E_{bn} = 7.85$ MeV.
• Comment worth adding: iron sits at the top of the curve and bismuth on its falling side, exactly as the graph predicts.

A copper coin (NCERT 13.3)

• $3.0$ g of ${}^{63}_{29}\text{Cu}$ ($m = 62.92960$ u): number of atoms $N = \dfrac{6.023\times10^{23}\times3.0}{63} = 2.868\times10^{22}$.
• Per nucleus: $\Delta m = 29(1.007825) + 34(1.008665) - 62.92960 = 0.591935$ u, so $E_b = 551.4$ MeV.
• Total energy to separate all the nucleons $= 2.868\times10^{22}\times551.4 \approx 1.58\times10^{25}$ MeV $\approx 2.5\times10^{12}$ J.

Standard results

• $E = mc^2$; $1\ \text{u} = 931.5\ \text{MeV}/c^2$; energy equivalent of 1 g is $9\times10^{13}$ J.
• $\Delta M = [Zm_p + (A-Z)m_n] - M$, or with atomic data $[Zm_H + (A-Z)m_n] - M_{\text{atom}}$.
• $E_b = \Delta M c^2 = \Delta M(\text{u})\times931.5$ MeV; $E_{bn} = E_b/A$.
• Curve: max $\approx 8.75$ MeV at $A = 56$; $\approx 8.0$ MeV flat for $30 < A < 170$; $7.6$ MeV at $A = 238$; low for $A < 30$ and $A > 170$.
• Constancy of $E_{bn}$ $\Rightarrow$ saturation $\Rightarrow$ short range of the nuclear force.
• Fusion of light nuclei and fission of heavy nuclei both RAISE $E_{bn}$, so both release energy.
• Representative $E_{bn}$: ${}^{14}\text{N}$ $7.48$ MeV, ${}^{56}\text{Fe}$ $8.79$ MeV, ${}^{209}\text{Bi}$ $7.85$ MeV, ${}^{4}\text{He}$ $7.07$ MeV.