NCERT Solutions Class 12 Physics Chapter 13 Nuclei

Atomic vs nuclear mass — the most common pitfall. Tables of nuclidic masses (including the NCERT data) give the atomic mass — nucleus plus all the bound electrons. If you mistakenly used the bare proton mass m_p = 1.007276 u instead of m_H = 1.007825 u, you'd forget the seven electrons on the right side of the equation, and your Δ m would be slightly off. Using m_H lets the electron mass cancel cleanly between both sides.

The 931.5 MeV/u shortcut. Whenever Δ m is expressed in atomic mass units, you do not need to plug in c² explicitly. Just multiply by 931.5 MeV per u. This single conversion saves an entire 1.6610^-27310⁸²/1.610^-13 calculation.

Where this energy comes from physically. The strong nuclear force binds nucleons together. To rip nitrogen-14 apart into 14 free particles, you would need to supply 104.7 MeV of energy. By Einstein's equivalence, that energy "lives" as a mass deficit — the assembled nucleus literally weighs less than its pieces.

Nuclear-stability curve. Plotting binding energy per nucleon E_b/A against mass number A gives the famous curve that rises sharply from hydrogen, peaks near ⁵⁶Fe at about 8.79 MeV/nucleon, then slowly falls for heavier elements. Nitrogen-14's value of 7.48 MeV/nucleon sits on the steep rising part — explaining why light nuclei fuse to release energy (climbing toward iron) and heavy nuclei fission to release energy (also climbing toward iron from the other side).

Real-world relevance. The same calculation applied to deuterium and tritium underlies the energy budget of fusion reactors like ITER; applied to uranium-235, it explains the ∼ 200 MeV released per fission in a power-plant reactor. Every "binding energy" you compute is, in principle, energy that could be released if the nucleus could be rearranged toward iron.

Why iron is the "peak" of nuclear stability. The E_b/A curve has its maximum near A ≈ 56 (specifically nickel-62 is the absolute peak, but iron-56 is almost tied and far more abundant). To the left of the peak (light nuclei), nuclei gain stability by fusing together to make heavier ones — this is what powers stars. To the right of the peak (heavy nuclei like uranium or bismuth), nuclei gain stability by fissioning into lighter ones. Iron is the "ash" of stellar nucleosynthesis: a star can extract energy from fusion only up to iron, after which fusing further actually costs energy.

Alternate method — direct multiplication. A slicker way to write the calculation: E_b = Z m_H + (A-Z)m_n - m_atom × 931.5 MeV. Bake the conversion factor into a single line so you don't lose track of intermediate Δ m values.

Common mistake — using atomic mass of hydrogen ion. Some tables provide m_p (mass of bare proton) instead of m_H (mass of hydrogen ATOM including the electron). If you use m_p but the table's m(⁵⁶₂₆Fe) includes 26 electrons, you'll be off by Z m_e — about 0.014 u for iron, which translates to a ∼ 13 MeV error.

Real-world relevance — nuclear reactors. Bismuth-209 was, for a long time, considered the heaviest stable nucleus (we now know it's very weakly radioactive). The fact that E_b/A drops off for heavy nuclei is what makes them susceptible to fission. In a uranium reactor, breaking ²³⁵U E_b/A ≈ 7.6 MeV into fragments near iron-56 E_b/A ≈ 8.8 MeV releases roughly 200 MeV per fission — the energy difference times 235 nucleons.

Why ordinary chemistry cannot tap this. Chemical bonds are formed and broken at energies of a few eV per atom — nuclear binding is a million times larger. A normal flame can't even slightly nudge a nucleus; you need bombarding particles with MeV-scale kinetic energy (cyclotrons, fission products) to disturb nuclei.

Comparison with macroscopic energies. 2.5× 10¹² J is the energy released by burning approximately 600 tonnes of TNT, or by burning about 60 tonnes of petrol. All trapped inside a 3-gram coin. This is the basic insight behind nuclear power and weapons — even partial unlocking of nuclear binding energy yields stupendous amounts of energy from tiny amounts of matter.

Common mistake — confusing total binding with reaction energy. The energy computed here is the cost to dismantle each nucleus into free nucleons. Real nuclear reactions (fission, fusion) rearrange nucleons from less-bound to more-bound configurations, releasing only the difference in binding energies — typically ∼ 1 MeV/nucleon, not the full ∼ 8.7 MeV/nucleon. So actual reactor energy yields are about ten times smaller than the naive total binding energy.

Order-of-magnitude shortcut. For any solid sample, the number of atoms is ∼ N_A M/M_mol and the per-nucleus binding energy is ∼ A× 8 MeV. Combining: total binding energy ∼ 8 MeVN_A M / m_u where m_u is the atomic mass unit. For 3 g of matter, this gives ∼ 10²⁵ MeV. The detailed calculation above lands on 1.6× 10²⁵ MeV — consistent.

Why nuclear density is constant. If R ∝ A^1/3, then volume V = 43π R³ ∝ A. Density ρ = nucleon mass × A/V is independent of A — every nucleus, light or heavy, has the same density: about 2.3× 10¹⁷ kg/m³. One teaspoon of nuclear matter would weigh nearly a billion tonnes. This is what makes neutron stars so extreme: they're essentially giant nuclei held together by gravity instead of the strong force.

Alternate method — log scaling. When ratios involve cube roots, taking logs is sometimes faster: log(R_Au/R_Ag) = 13log(197/107) = 13log(1.84) ≈ 13(0.265) ≈ 0.088. So R_Au/R_Ag ≈ 10^0.088 ≈ 1.22. Same answer.

Why protons don't blow the nucleus apart. Even though all those positively-charged protons repel each other electrostatically, the strong nuclear force between every pair of nucleons (proton-proton, proton-neutron, neutron-neutron) is far stronger at femtometre distances. It saturates with the nearest neighbours (short-range), giving rise to the nearly-constant nuclear density.

Common mistake — using Z instead of A. The radius formula uses A, not Z — adding neutrons makes the nucleus bigger just as much as adding protons. Confusing A and Z here is a classic JEE/NEET trap.

Why electron masses cancel. The left side of reaction (i) carries 1 + 1 = 2 electrons (one in each hydrogen atom); the right side also carries 1 + 1 = 2 electrons. Both sides include the same total electron mass, so it drops out of Δ m. The same is true for reaction (ii): 6 + 6 = 12 on both sides. This is why we can use atomic masses directly in Q-value calculations as long as the total Z is conserved (which it always is, since reactions don't create or destroy charge).

What endothermic vs exothermic means experimentally. Reaction (i) cannot proceed spontaneously — to make it happen, you must bombard a target tritium nucleus with a proton whose kinetic energy exceeds the 4.03 MeV "threshold" (plus a bit more in the lab frame because of momentum conservation). Reaction (ii) (carbon-carbon fusion) is exothermic — it's one of the energy-releasing steps deep inside massive stars approaching the supernova stage.

Stellar context. The fusion of ¹²C + ¹²C only happens at core temperatures of a billion kelvin or more — only in stars eight times more massive than the Sun. Lower-mass stars (like the Sun) stop at helium burning. So reaction (ii) is what allows heavy stars to build oxygen, neon, magnesium, silicon up to iron, populating the periodic table.

Common mistake — sign confusion. NCERT writes Q = m_A + m_b - m_C - m_dc² — reactants minus products. Some textbooks reverse the convention (products minus reactants). With the reversed convention you get Q_(i) = +4.03 MeV and Q_(ii) = -4.62 MeV — and you'd flip "exo" and "endo". Always check the sign convention printed at the top of the problem.

Why iron resists splitting. Iron-56 sits very close to the peak of the binding-energy-per-nucleon curve E_b/A ≈ 8.79 MeV. Aluminium-28 lies on the rising side of the same curve — it is less tightly bound per nucleon. Going from iron to two aluminiums is climbing down the stability hill, so it costs energy. Only nuclei beyond the peak (uranium, plutonium, thorium) lose mass by splitting.

Alternate method — binding-energy difference. Q = E_b(products) - E_b(reactant) . For iron-56, E_b ≈ 492 MeV. For Al-28, E_b ≈ 224 MeV. Two aluminiums give 2(224)=448 MeV. Difference: 448 - 492 = -44 MeV the exact value differs from the direct calculation because of rounding in E_b values. Either method gives Q < 0 — fission impossible.

Where fission DOES work. For ²³⁵U, splitting into two fragments near the middle of the periodic table (Kr-92 + Ba-141, for example) releases about 200 MeV per event — this is the principle behind every fission reactor in the world. The split moves the system from low E_b/A uranium, ∼ 7.6 MeV toward higher E_b/A Kr/Ba, ∼ 8.6 MeV.

Real-world significance — stellar end-states. Stars burn elements in increasing order, climbing the binding-energy curve all the way up to iron. Once the core is iron, fusion stops releasing energy. Without thermal pressure, gravity wins and the star collapses — leading to a supernova. This calculation, in miniature, is the very physics that ends massive stars' lives.

Why fission is so energy-dense. The energy per fission 180 MeV is millions of times larger than a typical chemical reaction (a few eV per molecule). Per gram: ∼ 7.3× 10¹⁰ J/g for Pu fission, versus ∼ 5× 10⁴ J/g for petrol combustion. Nuclear is roughly a million times more concentrated.

Alternative shortcut — energy per kilogram from E = mc². A useful sanity check: if all 1 kg were converted entirely to energy, you'd get mc² = (1)3× 10⁸² = 9× 10¹⁶ J. Our calculated 7.3× 10¹³ J is about 0.08% of that — meaning fission converts roughly 0.1% of the mass into energy. This is consistent with the fact that fission of ²³⁹Pu drops the per-nucleon binding energy from ∼ 7.5 MeV to ∼ 8.5 MeV, a 1 MeV-per-nucleon difference out of about 940 MeV/nucleon rest mass — about 0.1%.

Real-world relevance — reactors and bombs. 1 kg of Pu-239 is roughly the "critical mass" for a nuclear weapon (with implosion geometry). A reactor uses much more fuel because only a fraction is allowed to fission at any time, but the same physics applies. The Pu-239 produced in breeder reactors from ²³⁸U absorbing a neutron is the standard alternative to ²³⁵U — and its fission products are very similar to uranium's.

Common mistake — forgetting molar mass. A surprising number of students compute N = 1 kg × N_A and get 6× 10²⁶ atoms — too high by a factor of 239. Always divide by the molar mass first.

Why two deuterons per event matter. A very common error: students compute N = 6× 10²⁶ deuterons and multiply by 3.27 MeV directly. But the reaction equation shows TWO deuterons combining to release 3.27 MeV. Each event consumes two — so the number of events is N/2, not N. Forgetting this halves your answer.

Real-world relevance — fusion power. Deuterium is abundant in seawater (0.015% of all hydrogen is deuterium). 2 kg of deuterium can in principle be extracted from a few hundred cubic metres of seawater. The energy yield calculated here, 1.6× 10¹⁴ J, is enormous — but achieving controlled fusion requires temperatures of ∼ 10⁸ K to overcome the Coulomb barrier between the two positively-charged deuterons. Tokamaks like ITER are the engineering attempt.

Alternative reaction — D+T fusion. Real reactors use ²H + ³H → ⁴He + n + 17.6 MeV, which has a much higher cross-section than D+D at attainable temperatures. The 17.6 MeV per event is about 5× the D+D yield, but tritium has to be bred from lithium in the reactor blanket. The principle of energy calculation is identical to what we just did.

Common mistake — fusion vs fission energy density. Compared to Q13.7 Pu fission, ∼ 7× 10¹³ J per kg, D-D fusion gives ∼ 8× 10¹³ J per kg of deuterium. Similar order of magnitude. But fusion's fuel is essentially limitless (seawater), whereas fissile uranium is a finite mined resource.

Why this height matters. For two deuterons to fuse, they must approach close enough for the strong nuclear force range ∼ 1 fm to glue them together. To get that close, each must climb a Coulomb barrier roughly 0.36 MeV tall. Their kinetic energy must therefore be of comparable magnitude. At thermal equilibrium, KE= 32k_B T. Setting this equal to 0.36 MeV gives T ∼ 10⁹ K — a billion kelvin.

Quantum tunnelling rescues fusion. Stars only reach core temperatures of about 10⁷ K — a hundred times too cold to surmount the deuteron barrier classically. Fusion happens anyway because particles can tunnel through the barrier (a purely quantum effect). The tunnelling probability is exponentially small but nonzero. This was Gamow's great insight in the 1920s; it explains how the Sun (and tokamaks) work at "modest" temperatures.

Alternate units. Energies of order MeV are easy to express in joules: 1 MeV = 1.6× 10^-13 J, 1 keV = 1.6× 10^-16 J. Memorising these makes back-of-envelope nuclear estimates fast.

Common mistake — using the deuteron radius instead of the centre-to-centre distance. Students sometimes use r = 2.0 fm (the radius of one deuteron) in the denominator instead of r = 4.0 fm (the centre-to-centre distance when surfaces touch). That doubles the answer to ∼ 720 keV. Always picture the geometry: each sphere centre is one radius away from the contact point.

Real-world relevance — controlled fusion difficulty. This barrier-height calculation is exactly why fusion power is so hard. ITER is designed to run at ∼ 1.5× 10⁸ K — about ten times the core of the Sun — to provide enough deuterons and tritons in the Maxwell-Boltzmann tail to tunnel the barrier at a useful rate. The 360 keV figure is the energy scale that defines all of fusion engineering.

Why this matters physically — saturation of the strong force. If the strong force were like gravity or electromagnetism (long-range, unscreened), nuclear density would increase with A — every nucleon would feel every other one, packing them tighter. Instead, density is independent of A. This tells us the strong force is short-range: each nucleon only feels its immediate neighbours. Adding more nucleons doesn't compress the existing ones — it just makes the nucleus bigger. This is called "saturation."

Mind-bending comparison. Water has density 1000 kg/m³; lead has ∼ 11000 kg/m³; a white dwarf core, ∼ 10⁹ kg/m³. Nuclear matter at 2.3× 10¹⁷ kg/m³ is about 10¹⁴ times denser than water. One cubic millimetre of nuclear matter would weigh ∼ 2× 10⁸ kg — 200 thousand tonnes.

Neutron stars — macroscopic nuclei. When a massive star collapses, its electrons get crushed into protons forming neutrons. The remnant — a neutron star — is essentially a giant nucleus held together by gravity instead of the strong force, with mass roughly the Sun's compressed to a radius of about 10 km. Its density ∼ 10¹⁷ kg/m³ matches the nuclear value we just derived. The match is exact for a reason: at those densities, neutron-star matter is nuclear matter.

Alternate derivation — number density of nucleons. Equivalently, define number density n = A/V = A/43π R₀³ A = 1/43π R₀³ ≈ 0.14 nucleons/fm³. The same answer in different units. This is the universal "nuclear saturation density" used in heavy-ion collision physics and neutron-star modelling.

Real-world relevance. The constancy of nuclear density underpins the liquid-drop model of the nucleus (Bohr, Weizsäcker), the semi-empirical mass formula, and the analysis of giant-resonance excitations. Whenever you see "nuclear matter density," it is exactly the ∼ 2.3× 10¹⁷ kg/m³ computed above.