Chapter 13 of 14

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CBSE Class 12 Physics★ Unit VIII · Atoms and Nuclei, CBSE states no chapter-wise split

Nuclei

Nuclear composition and size, mass defect and binding energy, the binding energy per nucleon curve, nuclear forces, radioactivity, and energy from fission and fusion. Section 13.6 has been cut to a single paragraph, and the decay law, half-life and mean life survive only in the end-of-chapter symbol table. CBSE Class 12 Physics Chapter 13.

⏱ 10 min readMedium difficulty✓ Free · NCERT-aligned

Worksheet

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Mind Map

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Describe nuclear composition and distinguish isotopes, isobars and isotones
2Apply R = R_0 A^(1/3) and show that nuclear density is independent of mass number
3Compute mass defect and binding energy using atomic masses consistently
4Explain why binding energy per nucleon rather than total binding energy measures stability
5Interpret the binding energy curve and use it to explain fission and fusion
6Compute the Q value of a nuclear reaction and decide whether it is exothermic
7State the properties of the nuclear force and explain saturation
8Describe the three decay types and compute energy yields from fission and fusion

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Why this chapter matters

Binding energy per nucleon is the idea that explains both fission and fusion, and it appears in the paper every year. The mass defect calculation is among the most predictable numerical questions in the whole syllabus.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Atoms (Class 12 Ch 12): alpha-particle scattering established the existence and size of the nucleus that this chapter examines

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Dual Nature of Radiation and Matter (Class 12 Ch 11): the mass-energy relation and the electronvolt are used throughout

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Electric Charges and Fields (Class 12 Ch 1): Coulomb repulsion between protons is what the nuclear force must overcome, and it sets the fusion barrier

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Work, Energy and Power (Class 11 Ch 5): energy conservation underlies every Q value calculation

Nuclei

1. Check this before you revise anything

The "Additional Exercises" section has been removed from this chapter, as from all 14 chapters of the current Class 12 Physics book, leaving 10 questions.

Section 13.6 Radioactivity has been cut to a single paragraph. It now contains Becquerel's 1896 discovery and a three-line list stating that $α$ -decay emits a helium nucleus, $β$ -decay emits electrons or positrons, and $γ$ -decay emits high-energy photons. Then the chapter moves straight on to 13.7 Nuclear Energy.

The law of radioactive decay is gone. Searching the chapter returns zero hits for "law of radioactive", "exponential" and the form $N = N_{0} e^{- λ t}$ . There is no derivation, no graph and no worked example.

Yet the end-of-chapter symbol table still lists all of it. That table defines the decay constant $λ$ , the half-life $T_{1/2}$ ("time taken for the decay of one-half of the initial number of nuclei"), the mean life $τ$ ("time at which the number of nuclei has been reduced to $e^{- 1}$ of its initial value") and the activity $R$ in becquerel.

So four quantities are defined in a summary table that no section of the chapter teaches. This is the same pattern as the quality factor in Chapter 7.

The exercises are at least self-consistent. None of the ten questions requires the decay law, so the chapter can be worked through as printed. But any past paper or practice sheet asking for half-life calculations is drawing on material this edition no longer contains.

Also removed: the detailed treatment of $α$ -, $β$ - and $γ$ -decay with their Q-values and decay schemes, and the reactor moderator (zero hits). The curie as a unit returns zero hits.

Textbook section	Topic
13.1 to 13.2	Introduction; atomic masses and composition of the nucleus
13.3	Size of the nucleus
13.4	Mass-energy and nuclear binding energy
13.5	Nuclear force
13.6	Radioactivity — one paragraph
13.7	Nuclear energy: fission and fusion

2. Nuclear Composition and Size (Textbook 13.2 to 13.3)

A nucleus contains $Z$ protons and $A - Z$ neutrons, together called nucleons, where $Z$ is the atomic number and $A$ the mass number.

The three "iso" families, which are easy to confuse:

	Same	Example
Isotopes	Same $Z$ , different $A$	$^{35}$ Cl and $^{37}$ Cl
Isobars	Same $A$ , different $Z$	$^{3}$ H and $^{3}$ He
Isotones	Same number of neutrons	$^{3}$ H and $^{4}$ He

Atomic mass unit. One u is $\frac{1}{12}$ of the mass of a $^{12}$ C atom, and the energy equivalent is:

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

Nuclear size. Scattering experiments give:

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

The radius depends only on the mass number, not the charge. Because the cube root compresses differences, a gold nucleus is only about 23 per cent larger than a silver one despite nearly double the mass — Exercise 13.4.

Nuclear density is constant. Since $M \propto A$ and $V \propto R^{3} \propto A$ , the $A$ cancels:

$ρ = \frac{3 m}{4 π R _{0}^{3}} \approx 2.3 \times 1 0^{17} kg m^{- 3}$

This is about $1 0^{14}$ times the density of water and is the same for every nucleus, showing nuclear matter is essentially incompressible. That is Exercise 13.10.

3. Mass Defect and Binding Energy (Textbook 13.4)

The measured mass of a nucleus is always less than the sum of its parts. The shortfall is the mass defect:

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

By Einstein's mass-energy relation this missing mass appears as the binding energy:

$E_{b} = Δ m c^{2} = Δ m \times 931.5 MeV$

This is the energy that would be needed to pull the nucleus completely apart, and equally the energy released when it was assembled.

Use atomic masses consistently. The $m_{H}$ in the formula is the hydrogen atom, so the $Z$ electron masses it contributes cancel against those already included in the atomic mass of the nucleus. Mixing atomic and nuclear masses is the standard error.

Binding energy per nucleon is what measures stability, not the total:

$\frac{E _{b}}{A}$

Exercise 13.2 makes the point directly. Bismuth-209 has a far larger total binding energy than iron-56, 1640 MeV against 492 MeV, yet iron is the more tightly bound nucleus at 8.79 MeV per nucleon against 7.85 MeV.

The curve of binding energy per nucleon is the single most important graph in the chapter:

It rises steeply for light nuclei.
It is nearly flat at about 8.0 MeV between $A = 30$ and $A = 170$ .
It peaks near $A = 56$ , at iron, around 8.8 MeV.
It falls slowly for heavy nuclei, reaching about 7.6 MeV at uranium.

Everything about nuclear energy follows from this shape. Any process moving nucleons towards the peak releases energy. Heavy nuclei beyond the peak release energy by splitting; light nuclei below it release energy by joining.

Exercise 13.6 is the cleanest illustration: splitting iron-56 into two aluminium-28 nuclei gives $Q = - 26.9$ MeV. It is negative, so the process is impossible, precisely because iron already sits at the peak and any split moves downhill.

4. Nuclear Force (Textbook 13.5)

A nucleus packs many positively charged protons into $1 0^{- 15}$ m, where Coulomb repulsion is enormous. Something far stronger must hold it together.

Properties of the nuclear force:

Strongest of the known forces in its range, around 100 times the electrostatic force.
Very short ranged, effective only to a few femtometres, and dropping to nothing beyond.
Charge independent: the force between two protons, two neutrons, or a proton and a neutron is essentially the same.
Saturated: each nucleon interacts only with its immediate neighbours, not with every other nucleon.
Strongly repulsive at very short distances, below about 0.8 fm, which stops the nucleus collapsing.

Saturation explains the flat curve. If every nucleon attracted every other, binding energy would grow as $A^{2}$ . That it grows roughly as $A$ instead — leaving $E_{b} / A$ nearly constant — is direct evidence that each nucleon binds only to its neighbours.

Unlike Coulomb's law or gravitation, the nuclear force has no simple mathematical form.

5. Radioactivity and Nuclear Energy (Textbook 13.6 to 13.7)

The three decay types, which is all section 13.6 now retains:

Decay	Emits	Change
$α$	A helium nucleus $_{2}^{4}$ He	$Z \to Z - 2$ , $A \to A - 4$
$β$	An electron or a positron	$Z$ changes by 1, $A$ unchanged
$γ$	High-energy photons, hundreds of keV upward	Neither changes

$γ$ -emission follows $α$ or $β$ decay, carrying away energy from a nucleus left in an excited state.

Fission. A heavy nucleus splits into two intermediate fragments, releasing about 200 MeV per event — some $1 0^{7}$ times a chemical reaction. Exercise 13.7 works out $4.54 \times 1 0^{26}$ MeV from a single kilogram of plutonium-239.

Fusion. Light nuclei join to form a heavier one, releasing energy because the product lies higher on the binding curve. This powers the Sun.

Why fusion is hard. Both nuclei are positively charged and must be forced close enough for the short-range nuclear force to act. Exercise 13.9 computes the Coulomb barrier for two deuterons at contact as $0.36$ MeV, which corresponds to temperatures near $1 0^{9}$ K. That is why fusion needs stellar conditions while fission proceeds at ordinary temperatures.

The energy density is extraordinary. Exercise 13.8 finds that 2 kg of deuterium could keep a 100 W lamp glowing for about fifty thousand years.

Summary

A nucleus has $Z$ protons and $A - Z$ neutrons; isotopes share $Z$ , isobars share $A$ , isotones share the neutron number.
$1 u = 931.5$ MeV/c $^{2}$ .
$R = R_{0} A^{1/3}$ with $R_{0} = 1.2 \times 1 0^{- 15}$ m — radius depends on mass number only, not charge.
Nuclear density $ρ = \frac{3 m}{4 π R _{0}^{3}} \approx 2.3 \times 1 0^{17}$ kg m $^{- 3}$ , the same for every nucleus.
Mass defect $Δ m = [Z m_{H} + (A - Z) m_{n}] - M$ , and $E_{b} = Δ m \times 931.5$ MeV.
Use atomic masses throughout, since the electron masses cancel.
Binding energy per nucleon, not total binding energy, measures stability.
Bi-209 has more total binding energy than Fe-56, but Fe-56 is more tightly bound per nucleon.
The curve is flat near 8.0 MeV for $30 < A < 170$ and peaks at iron, $A \approx 56$ , near 8.8 MeV.
Moving towards the peak releases energy: heavy nuclei by fission, light nuclei by fusion.
Splitting iron gives $Q = - 26.9$ MeV — negative, hence impossible, since iron is already at the peak.
The nuclear force is the strongest known at short range, short-ranged, charge-independent, saturated, and repulsive below 0.8 fm.
Saturation is why $E_{b} / A$ is nearly constant rather than growing with $A$ .
$α$ -decay changes $Z$ by $- 2$ and $A$ by $- 4$ ; $β$ -decay changes $Z$ by 1 with $A$ fixed; $γ$ -decay changes neither.
Fission releases about 200 MeV per event, roughly $1 0^{7}$ times a chemical reaction.
Fusion needs about $1 0^{9}$ K, because the Coulomb barrier for two deuterons is $0.36$ MeV.
A positive $Q$ means exothermic and a negative $Q$ means the reaction cannot proceed spontaneously.
Section 13.6 is now one paragraph: the decay law, half-life, mean life and activity are defined only in the end-of-chapter symbol table, with no section teaching them.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Nuclear composition

A nucleus has Z protons and A - Z neutrons

Isotopes share Z, isobars share A, isotones share the neutron number

Atomic mass unit

1 u = 1/12 the mass of a carbon-12 atom = 931.5 MeV/c squared

The conversion factor used in every binding energy calculation

Nuclear radius

R = R_0 A^(1/3), with R_0 = 1.2 x 10^-15 m

Depends only on the mass number, not on the charge

Nuclear density

rho = 3m/(4 pi R_0 cubed), about 2.3 x 10^17 kg per cubic metre

A cancels completely, so every nucleus has the same density, about 10^14 times that of water

Mass defect

Delta m = [Z m_H + (A - Z) m_n] - M

Use the atomic mass of hydrogen so the electron masses cancel against those in the atomic mass

Binding energy

E_b = Delta m x 931.5 MeV

The energy needed to separate the nucleus completely into free nucleons

Binding energy per nucleon

E_b / A

This, not the total, measures stability: Bi-209 has more total binding energy than Fe-56 yet is less tightly bound

Shape of the binding energy curve

Flat near 8.0 MeV for A between 30 and 170, peaking near A = 56 at about 8.8 MeV

Falls to about 7.6 MeV at uranium, which is why heavy nuclei can release energy by splitting

Q value of a reaction

Q = (sum of initial masses - sum of final masses) x 931.5 MeV

Positive Q means exothermic; negative Q means the reaction cannot proceed spontaneously

Properties of the nuclear force

Strongest at short range, very short ranged, charge independent, saturated, repulsive below about 0.8 fm

Saturation is why binding energy grows as A rather than A squared

Alpha decay

Emits a helium nucleus: Z decreases by 2 and A decreases by 4

One of only three decay types the current chapter retains

Beta decay

Emits an electron or a positron: Z changes by 1 while A is unchanged

The nucleon number is conserved because a neutron converts to a proton or the reverse

Gamma decay

Emits high energy photons; neither Z nor A changes

Follows alpha or beta decay, carrying away energy from an excited nucleus

Energy from fission

About 200 MeV per event for a heavy nucleus

Roughly 10^7 times the energy of a chemical reaction per atom

Coulomb barrier for fusion

U = k e squared / r, with r taken as the sum of the two radii

For two deuterons in contact this is 0.36 MeV, needing temperatures near 10^9 K

⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Judging stability by total binding energy instead of binding energy per nucleon

✓ Bismuth-209 has 1640 MeV of total binding energy against iron-56's 492 MeV, yet iron is the more stable nucleus at 8.79 MeV per nucleon against 7.85. Always divide by A.

WATCH OUT

✗ Mixing atomic and nuclear masses in the mass defect formula

✓ Use atomic masses throughout. The Z electron masses contributed by m_H cancel against those already contained in the atomic mass of the nucleus.

WATCH OUT

✗ Assuming any heavy nucleus can be split to release energy

✓ Only nuclei beyond the peak of the curve. Exercise 13.6 shows that splitting iron-56 gives Q = -26.9 MeV, so it is impossible, because iron already sits at the peak.

WATCH OUT

✗ Taking the nuclear radius to depend on the atomic number

✓ R = R_0 A^(1/3) contains only the mass number. Two isobars of very different charge have essentially the same radius.

WATCH OUT

✗ Forgetting that each fusion reaction consumes two nuclei

✓ In Exercise 13.8 the number of reactions is half the number of deuterium atoms, since two deuterons fuse each time. Missing this doubles the answer.

WATCH OUT

✗ Using the radius rather than twice the radius for two touching spheres

✓ When two deuterons of radius 2.0 fm just touch, their centres are 4.0 fm apart. Using 2.0 fm doubles the computed barrier height.

WATCH OUT

✗ Expecting nuclear density to vary between elements

✓ It does not. Mass goes as A and volume also goes as A, so the ratio is a constant of about 2.3 x 10^17 kg per cubic metre for every nucleus.

WATCH OUT

✗ Preparing the law of radioactive decay from this chapter

✓ It is not there. Zero hits are returned for the exponential decay law, and section 13.6 has been reduced to a single paragraph naming the three decay types.

WATCH OUT

✗ Assuming half-life and mean life are taught because they appear in the symbol table

✓ The end-of-chapter table defines the decay constant, half-life, mean life and activity, but no section of the chapter develops them. None of the ten exercises requires them either.

WATCH OUT

✗ Thinking the nuclear force obeys an inverse square law

✓ It has no simple mathematical form, unlike Coulomb's law or gravitation. It is short ranged, charge independent, saturated, and becomes repulsive at very small separations.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Binding energies of nitrogen, iron and bismuth, the energy to disassemble a copper coin, the ratio of nuclear radii, Q values of fusion reactions, the impossibility of iron fission, energy from plutonium fission and deuterium fusion, the Coulomb barrier between deuterons, and the constancy of nuclear density

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Nuclei?

9 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

9 questions~6 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•A nucleus has Z protons and A - Z neutrons; isotopes share Z, isobars share A, isotones share the neutron number
•1 u = 931.5 MeV/c squared
•R = R_0 A^(1/3) with R_0 = 1.2 x 10^-15 m; radius depends on mass number only
•Nuclear density is about 2.3 x 10^17 kg per cubic metre and is the same for every nucleus
•Mass defect = [Z m_H + (A - Z) m_n] - M, and binding energy = mass defect x 931.5 MeV
•Use atomic masses throughout so the electron masses cancel
•Binding energy PER NUCLEON measures stability, not the total
•Bi-209 has more total binding energy than Fe-56 but is less tightly bound per nucleon
•The curve is flat near 8.0 MeV for A between 30 and 170 and peaks at iron near 8.8 MeV
•Moving towards the peak releases energy: heavy nuclei by fission, light nuclei by fusion
•Splitting iron gives Q = -26.9 MeV, so it is impossible
•The nuclear force is strongest at short range, short ranged, charge independent, saturated and repulsive below 0.8 fm
•Saturation is why binding energy grows as A rather than A squared
•Alpha decay: Z falls by 2 and A by 4. Beta decay: Z changes by 1, A fixed. Gamma decay: neither changes
•Fission releases about 200 MeV per event, around 10^7 times a chemical reaction
•The Coulomb barrier for two deuterons is 0.36 MeV, needing about 10^9 K for fusion
•Positive Q means exothermic; negative Q means the reaction cannot proceed spontaneously
•Section 13.6 is now a single paragraph; the decay law, half-life, mean life and activity appear only in the symbol table
•The Additional Exercises block has been removed, leaving Exercises 13.1 to 13.10

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit VIII: Atoms and Nuclei, no chapter-wise split published by CBSE

Question type	Marks each	Typical count	What it tests
Nuclear Radius and Density	2-3	1	The cube-root relation, ratios of radii, and the constancy of nuclear density
Mass Defect and Binding Energy, Binding Energy Curve and Stability	3-4	1	Mass defect computation, binding energy per nucleon, and the shape of the curve
Q Value of a Nuclear Reaction, Energy from Fission and Fusion, Coulomb Barrier for Fusion, and Alpha, Beta and Gamma Decay	3-5	1	Q values and their sign, energy yields, the fusion barrier, and decay types

Prep strategy

Use atomic masses consistently so the electron masses cancel automatically
Always divide binding energy by A before comparing two nuclei
Sketch the binding energy curve and mark the iron peak when explaining fission or fusion
Check the sign of Q and state explicitly whether the reaction is exothermic
Remember that two nuclei are consumed per fusion event when counting reactions
Do not revise the decay law, half-life or mean life from this chapter, since no section teaches them

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Nuclear power generation

Controlled fission of uranium-235 heats water to drive turbines, extracting millions of times more energy per kilogram than any chemical fuel.

Stellar energy

The Sun shines by fusing hydrogen into helium, which is why its output has remained steady for billions of years on a finite amount of fuel.

Medical imaging and therapy

Radioactive tracers reveal metabolic activity in PET scans, and targeted radiation destroys tumour tissue.

Radiocarbon dating

The steady decay of carbon-14 in once-living material dates archaeological finds, relying on the decay law this edition no longer teaches.

Smoke detectors

A small americium source ionises air in a chamber, and smoke entering interrupts the resulting current to trigger the alarm.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write the mass defect as an explicit labelled line before converting to energy

State which masses are atomic and confirm the electron cancellation

Divide by A before comparing any two nuclei for stability

Sketch and label the binding energy curve for any fission or fusion explanation

Give the sign of Q and name the reaction exothermic or endothermic in words

Halve the atom count when counting fusion reactions, since two nuclei fuse each time

Use twice the radius for the separation when two nuclei just touch

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

The semi-empirical mass formula, which reproduces the binding energy curve from volume, surface, Coulomb, asymmetry and pairing terms

STRETCH

The law of radioactive decay and its derivation, along with half-life, mean life and radioactive equilibrium, all removed from this edition

STRETCH

The liquid drop and shell models of the nucleus, and the origin of the magic numbers

STRETCH

Neutrino physics and the puzzle of the continuous beta spectrum that led Pauli to predict the neutrino

🚀

JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainTotal against per-nucleon binding energyComparing nuclear stability

Compute the binding energy and binding energy per nucleon of $_{26}^{56}$ Fe and $_{83}^{209}$ Bi, given $m (^{56} Fe) = 55.934939$ u and $m (^{209} Bi) = 208.980388$ u. Which nucleus is more stable?

Stuck? Show the approach

Find each mass defect separately, convert to energy, then divide by the mass number before comparing.

Show the full solution

For iron, $Δ m = 26 (1.007825) + 30 (1.008665) - 55.934939 = 0.528461$ u, so $E_{b} = 492.3$ MeV and $E_{b} / A = 8.79$ MeV. For bismuth, $Δ m = 83 (1.007825) + 126 (1.008665) - 208.980388 = 1.760877$ u, so $E_{b} = 1640.3$ MeV and $E_{b} / A = 7.85$ MeV. Bismuth has more than three times the total binding energy, but iron has the larger binding energy per nucleon, so iron is the more stable nucleus. Iron sits essentially at the peak of the binding energy curve, which is why it is the endpoint of both fission and fusion.

Answer: Fe-56: 492.3 MeV, 8.79 MeV per nucleon. Bi-209: 1640.3 MeV, 7.85 MeV per nucleon. Iron is more stable

The trap

Choosing bismuth because its total binding energy is larger. Stability is measured per nucleon.

JEE MainWhy iron cannot fissionQ value and the binding energy curve

Determine whether $_{26}^{56}$ Fe can split into two $_{13}^{28}$ Al nuclei, given $m (^{56} Fe) = 55.93494$ u and $m (^{28} Al) = 27.98191$ u.

Stuck? Show the approach

Compute Q from the mass difference and interpret its sign using the shape of the binding energy curve.

Show the full solution

$Q = [55.93494 - 2 (27.98191)] \times 931.5 = [55.93494 - 55.96382] \times 931.5 = (- 0.02888) \times 931.5 = - 26.9$ MeV. The Q value is negative, so the process absorbs energy rather than releasing it and cannot occur spontaneously. The reason is structural: iron-56 lies at the peak of the binding energy per nucleon curve at about 8.8 MeV, while aluminium-28 lies lower on the rising branch. Splitting iron therefore moves nucleons to a less tightly bound state, which costs energy. Fission releases energy only for nuclei beyond the peak, such as uranium.

Answer: Q = -26.9 MeV, so the fission is not energetically possible

The trap

Assuming any nucleus can be split for energy. Only those beyond the binding energy peak release energy on splitting.

JEE AdvancedNuclear density is universalDeriving density from the radius relation

Using $R = R_{0} A^{1/3}$ , show that nuclear matter density is independent of the mass number, and estimate its value.

Stuck? Show the approach

Express both the nuclear mass and the nuclear volume in terms of A, then divide and see what survives.

Show the full solution

A nucleus of mass number $A$ has mass $M \approx A m$ where $m$ is the nucleon mass. Its volume is $V = \frac{4}{3} π R^{3} = \frac{4}{3} π R_{0}^{3} A$ , since cubing the cube root returns $A$ exactly. Dividing, $ρ = \frac{A m}{\frac{4}{3} π R _{0}^{3} A} = \frac{3 m}{4 π R _{0}^{3}}$ , in which $A$ has cancelled completely. Substituting $m = 1.67 \times 1 0^{- 27}$ kg and $R_{0} = 1.2 \times 1 0^{- 15}$ m gives $ρ \approx 2.3 \times 1 0^{17}$ kg m $^{- 3}$ , about $1 0^{14}$ times the density of water. Every nucleus has this same density, which shows nuclear matter is incompressible and that nucleons pack as tightly as the nuclear force permits.

Answer: rho = 3m/(4 pi R_0 cubed), independent of A, and about 2.3 x 10^17 kg per cubic metre

The trap

Forgetting that cubing A^(1/3) gives A exactly, which is what makes the cancellation work.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

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Because total binding energy naturally grows with the number of nucleons, so a large nucleus will always have a large total simply by having more parts. Dividing by the mass number gives the average energy holding each individual nucleon in place, which is what actually determines how hard the nucleus is to break up. Exercise 13.2 shows this clearly: bismuth-209 has over three times the total binding energy of iron-56, yet iron is the more stable nucleus because each of its nucleons is bound by 8.79 MeV against bismuth's 7.85 MeV.

Because energy was released when the nucleons came together, and by Einstein's mass-energy relation that released energy corresponds to a loss of mass. The missing mass, called the mass defect, multiplied by c squared gives exactly the binding energy of the nucleus. To pull the nucleus apart again you would have to supply that same energy, which would restore the missing mass. The effect is far too small to notice in chemical reactions but is easily measurable at nuclear energies.

Because both move nucleons towards the peak of the binding energy per nucleon curve, which lies near iron at mass number 56. Heavy nuclei such as uranium sit beyond the peak at about 7.6 MeV per nucleon, so splitting them produces fragments closer to the peak with higher binding energy per nucleon, and the difference is released. Light nuclei such as hydrogen sit below the peak, so joining them also increases binding energy per nucleon. Iron itself is at the top, which is why Exercise 13.6 finds that splitting iron would absorb energy rather than release it.

Because the nuclear volume grows in exact proportion to the number of nucleons. The radius goes as the cube root of the mass number, so cubing it to get the volume returns the mass number exactly. Mass is also proportional to the mass number, so when you divide mass by volume the mass number cancels completely, leaving a constant that depends only on the nucleon mass and R nought. The physical meaning is that nucleons are packed at a fixed spacing, which follows from the nuclear force being short ranged and saturated.

Because two nuclei approaching each other are both positively charged and repel strongly, and they must be brought within a few femtometres before the short-range nuclear force can take over. Exercise 13.9 computes this Coulomb barrier for two deuterons as 0.36 MeV, which corresponds to temperatures of order 10 to the 9 kelvin. Fission has no such barrier, because a neutron is uncharged and can simply drift into a heavy nucleus and destabilise it, which is why fission reactors operate at ordinary temperatures while fusion requires stellar conditions.

No. Section 13.6 on radioactivity has been reduced to a single paragraph containing Becquerel's discovery and a three-line list of the alpha, beta and gamma decay types. Searching the chapter returns zero hits for the exponential decay law, and there is no derivation, graph or worked example. Curiously the end-of-chapter symbol table still defines the decay constant, the half-life, the mean life and the activity in becquerel, so four quantities are listed that no section teaches. None of the ten exercises requires them, so the chapter is workable as printed, but older practice papers will assume them.

It means each nucleon interacts only with its immediate neighbours rather than with every other nucleon in the nucleus. If every nucleon attracted every other one, the number of interacting pairs would grow as the square of the mass number and the binding energy would grow as A squared, making binding energy per nucleon rise steadily with size. What is actually observed is a nearly flat curve at about 8 MeV per nucleon across a wide range of mass numbers, which means binding energy grows roughly in proportion to A. That is direct evidence for saturation.

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Electrostatic Potential and Capacitance

Potential as a scalar, potential due to point charges and dipoles, equipotential surfaces, potential energy, conductors and dielectrics, and capacitors in series and parallel with their stored energy. The Additional Exercises block has been removed, leaving 11 questions. CBSE Class 12 Physics Chapter 2.

Current Electricity

Drift velocity, Ohm's law and resistivity, the temperature dependence of resistance, internal resistance and terminal voltage, and Kirchhoff's rules applied to networks including the Wheatstone bridge. The Additional Exercises block has been removed, leaving 9 questions. CBSE Class 12 Physics Chapter 3.

Moving Charges and Magnetism

The Lorentz force, circular and helical motion in a magnetic field, the Biot-Savart law, Ampere's circuital law, the solenoid, the force between parallel currents that defines the ampere, and the torque on a current loop. Two topics the chapter's own text still promises — the cyclotron and the toroid — have been removed from it. CBSE Class 12 Physics Chapter 4.

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