Skip to chapter text
PhysicsCh 13 Nuclei

Nuclei: Class 12 Physics · Chapter 13

Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 90 min
  • 16 practice questions
  • 29 MCQs

What you will learn in Nuclei

  • Describe the make-up of a nucleus with Z, N and A, use the atomic mass unit, and classify nuclides as isotopes, isobars or isotones
  • Explain how scattering experiments set the size of a nucleus, use R = R0 A^(1/3) and show that nuclear density is the same for all nuclei
  • Apply E = mc² to find mass defects, binding energies and binding energy per nucleon from atomic masses
  • Interpret the binding-energy curve and use it to explain saturation, fission and fusion
  • State the main properties of the nuclear force and read its potential-energy curve
  • Identify alpha, beta and gamma decay and the change each makes to Z and A
  • Calculate Q-values for nuclear reactions and compare nuclear and chemical energy
  • Explain how fission and thermonuclear fusion release energy, and why fusion needs very high temperatures

Original companion text aligned to NCERT Physics Class 12, Chapter 13, pp. 306–322. Not affiliated with or endorsed by NCERT.

13.1 Introduction

The sunlight warming your face left the Sun's core as energy set free when hydrogen nuclei joined to make helium. Some of the electricity in your home may come from Kudankulam or Tarapur, where uranium nuclei split. Both kinds of energy come from the nucleus (नाभिक), the tiny heart of the atom that the previous chapter uncovered with Rutherford's alpha-particle experiments.

Just how tiny is it? A typical atom is about 10−10 m across, while its nucleus is about 10−14 m: ten thousand times smaller in radius. Volume goes as the cube of the radius, so the nucleus fills only about (10−4)3=10−12 of the atom's volume. Picture an atom blown up to the size of a cricket stadium: the nucleus would be a grape sitting on the centre spot, and the electrons would roam the stands. Yet that grape carries almost all the mass. In a hydrogen atom the electron accounts for only 0.05 % of the mass; in heavier atoms the electrons' share is smaller still, so in every atom the nucleus holds over 99.9 % of the total mass.

3/3A nitrogen-14 nucleus holds 7 protons (+) and 7 neutrons (n).

Fig. 13.1Zooming in by a factor of ten thousand turns the dot at the centre of an atom into a nucleus made of protons and neutrons.NCERT p. 306

Step 1 shows an atom as a large dashed sky-blue cloud with three small electrons and a single dark dot at the centre, labelled as the nucleus, about ten to the minus ten metres across. Step 2 zooms in: the cloud swells and fades away, and the dot grows into a round cluster of fourteen balls, now labelled nucleus, about ten to the minus fourteen metres across. Step 3 marks each ball: seven protons carry a plus sign and seven neutrons the letter n, with a key and the label nitrogen-14, seven protons plus seven neutrons.

This chapter asks four questions about that grape. What is it made of, and how do we write down its make-up? How big is it, and how dense? Why does it hold together at all, when its protons repel one another? And how can rearranging nuclei release energy on the scale of power stations and stars? Along the way we meet radioactivity, fission and fusion.

Quick check
1 of 2
The radius of a nucleus is about 10−4 times the radius of its atom. Roughly what fraction of the atom's volume does the nucleus occupy?
Show answer

Answer (C) 10−12

Volume scales as radius cubed: (10−4)3=10−12. Choosing 10−4 forgets that a sphere's volume depends on R3, not R.

13.2 Atomic masses and composition of nucleus

A kilogram is a clumsy unit for atoms: one carbon atom has a mass of about 2×10−26 kg. Physicists and chemists instead use the atomic mass unit (symbol u), defined as one-twelfth of the mass of a single atom of carbon-12. We can find its value in kilograms from the mole. By definition of the mass unit, one mole of carbon-12 atoms has a mass of 12 g, and a mole holds Avogadro's number NA=6.022×1023 of atoms. So one carbon-12 atom has a mass of 0.012 kg/NA=1.9926×10−26 kg, and one-twelfth of that is the atomic mass unit.

Atomic mass unit
1 u=m(12C)12=1.660539×10−27 kg
(13.1)
Definition
Atomic mass unit (u)

One-twelfth of the mass of one carbon-12 atom: 1 u=1.6605×10−27 kg.

ExampleA hydrogen atom has a mass of 1.007825 u.

Measured in u, the masses of atoms come out close to whole numbers, as if every atom were built from blocks of about the mass of a hydrogen atom. Some elements break the pattern badly: copper comes out at 63.55 u and magnesium at 24.31 u. The explanation came from the mass spectrometer, which sorts ions by their charge-to-mass ratio and so weighs individual atoms. It showed that most elements are mixtures of atoms with the same chemistry but different masses. These are called isotopes (समस्थानिक), from Greek words meaning same place, because they share one place in the periodic table. The atomic mass printed in a periodic table is an average of the isotope masses, weighted by how common each isotope is (its relative abundance).

Example

Why boron is not 10 or 11

Natural boron is 19.9 % boron-10 (atomic mass 10.0129 u) and 80.1 % boron-11 (atomic mass 11.0093 u). Find the average atomic mass of boron.

Try it yourself first. Then reveal the solution one step at a time.

Strip the single electron from an ordinary hydrogen atom and what is left is a bare proton. So the proton's mass follows from two measured numbers, the mass of the hydrogen atom (1.007825 u) and the mass of the electron, me=0.000549 u:

Mass of the proton
mp=1.007825 u−0.000549 u=1.007276 u=1.67262×10−27 kg
(13.2)

Ordinary hydrogen accounts for 99.985 % of the hydrogen in nature. Almost all the rest is deuterium (atomic mass 2.014102 u), found in every sample of water. A third isotope, tritium (3.016049 u), is unstable, with half of any sample decaying in about 12 years, so the tritium used in laboratories is manufactured.

How many protons does a nucleus hold? Count charges. A neutral atom of atomic number Z has Z electrons around it, carrying −Ze in all, so its nucleus must carry +Ze. Each proton carries +e, and so the nucleus contains exactly Z protons. Could some electrons be hiding inside the nucleus, cancelling the charge of extra protons? Physicists once suspected so, but quantum mechanics forbids it: an electron squeezed into a region a few femtometres across would have far more energy than any nucleus could hold on to.

Discovery of neutron

The hydrogen isotopes posed a puzzle. All three nuclei have charge +e, so each has just one proton, yet their masses stand roughly in the ratio 1 : 2 : 3. Deuterium and tritium nuclei must therefore carry extra neutral matter, about one and two proton masses of it. In 1932 James Chadwick identified what that neutral matter is (Fig. 13.2).

4/4Energy and momentum balance only if the rays are neutral particles with nearly the mass of a proton.

Fig. 13.2Alpha particles striking beryllium release a neutral radiation that knocks fast protons out of paraffin wax; working back from those protons revealed the neutron.

A row of four objects from left to right: a box labelled alpha source, a thin purple slab labelled beryllium, a thicker pale-yellow slab labelled paraffin, and a box labelled counter with a needle. Step 1: an arrow carries alpha particles from the source to the beryllium. Step 2: a dashed arrow labelled neutral rays, not bent by E or B fields, runs from the beryllium to the paraffin. Step 3: a solid arrow carries fast protons, shown as a ball with a plus sign, from the paraffin to the counter, and the counter's needle swings. Step 4: a highlighted card shows a ball marked n with the words neutral, mass about equal to a proton.

Earlier experimenters had seen that beryllium bombarded with alpha particles gives off a very penetrating, uncharged radiation, and that this radiation can kick protons out of paraffin wax at high speed. They took it to be gamma rays. Chadwick showed that this cannot be right. For a gamma-ray photon to give a proton that much kinetic energy, it would need several times more energy than the alpha-beryllium collision could supply. A neutral particle about as heavy as the proton fits naturally: a moving ball transfers its energy best to a target ball of the same mass, as in a head-on collision of billiard balls. Applying conservation of energy and momentum to protons and to nitrogen nuclei knocked on by the rays, Chadwick found the new particle's mass to be very close to the proton's. He called it the neutron, and received the 1935 Nobel Prize in Physics for the discovery. Its mass is now known very precisely:

Mass of the neutron
mn=1.008665 u=1.674927×10−27 kg
(13.3)

The neutron is slightly heavier than the proton. A free neutron does not last: it turns into a proton, an electron and an antineutrino with a mean life of about 880 s, roughly a quarter of an hour. Bound inside a stable nucleus, the same neutron can last for ever.

We can now describe any nucleus by counting its particles. Three whole numbers do the job: Z counts the protons (it is the atomic number of the element), N counts the neutrons (the neutron number), and the mass number A counts both together:

Mass number
A=Z+N
(13.4)
where
Z
atomic number, the number of protons
N
neutron number, the number of neutrons
A
mass number, the total number of nucleons
Definition
Nucleon

A proton or a neutron, the two kinds of particle that make up a nucleus. The mass number A counts the nucleons.

A nuclear species, or nuclide, is written XZAX2Z2AX, where X is the chemical symbol. For example, X82208X2822208Pb is the most common nucleus of lead: 208 nucleons, of which 82 are protons and 208−82=126 are neutrons. Since the symbol already fixes Z, people often drop the subscript and say lead-208 or write Pb-208.

Isotopes now have a simple meaning: nuclei with the same Z but different N. Carbon-12, carbon-13 and carbon-14 all have 6 protons, with 6, 7 and 8 neutrons. Their atoms have identical electron arrangements and so the same chemistry, which is why chemists could not tell them apart. Tin holds the record, with ten stable isotopes. Two more families are useful. Isobars have the same mass number A: carbon-14, nitrogen-14 and oxygen-14 are isobars with A=14. Isotones have the same neutron number N: carbon-13, nitrogen-14 and oxygen-15 each have 7 neutrons. A chart with N along one axis and Z along the other shows all three families at a glance (Fig. 13.3).

3/4Isobars lie on a diagonal: same A, here 14 (C-14, N-14, O-14).

Fig. 13.3On an N–Z chart, isotopes share a row, isotones share a column and isobars lie along a diagonal.NCERT p. 309

A grid of nine rounded cells, with neutron number N equal to 6, 7 and 8 across the bottom and atomic number Z equal to 6, 7 and 8 up the side. The cells are labelled C-12, C-13, C-14 in the bottom row, N-13, N-14, N-15 in the middle row and O-14, O-15, O-16 in the top row; stable nuclides have solid borders and radioactive ones dashed borders, with a key below. Step 1 shows the bare chart and the rule A equals Z plus N. Step 2 shades the bottom row in green: isotopes, same Z, different N. Step 3 shades the diagonal from O-14 through N-14 to C-14 in orange: isobars, same A equal to 14. Step 4 shades the middle column from C-13 through N-14 to O-15 in blue: isotones, same N equal to 7.

InteractiveNuclide chart explorer

A chart of the light nuclides from the free neutron up to calcium, with neutron number N to the right and proton number Z upwards. Stable nuclides are solid ink-coloured squares and selected radioactive ones are pale orange squares, with a dashed N equals Z line. Buttons change Z and N, or you can tap a square. Cards show the nuclide's name, its mass number and its radius from R equals 1.2 A to the one-third femtometres. A status line gives the numbers of protons and neutrons, the ratio N over Z, whether the nucleus is stable or how it decays, and lists its isotopes, isobars or isotones when one of those highlight buttons is pressed.

Fig. 13.4Step through the light nuclei by changing Z and N, and highlight the isotopes, isobars or isotones of the one you pick.
Quick check
1 of 4
The iron nucleus X2656X226256Fe contains protons and neutrons.
Show answer

Answer (1) 26, (2) 30

Z=26 protons; N=A−Z=56−26=30 neutrons.

13.3 Size of the nucleus

How do you measure something 10−14 m across? You cannot look at it, so you throw things at it and watch how they bounce. In the Geiger–Marsden experiment of the previous chapter, alpha particles fired at gold foil were turned back by the electric repulsion of the gold nucleus. Rutherford's formula, which assumes nothing but Coulomb's law between two point charges, matched the results. That success already tells us something about size: the alphas never came close enough to feel anything other than electric repulsion.

In a head-on approach, the alpha particle slows down as its kinetic energy K turns into electric potential energy. It stops for an instant at the distance of closest approach d, where all of K has become potential energy, and then turns back. With charge 2e on the alpha particle and Ze on the nucleus,

Distance of closest approach
K=14πε0(2e)(Ze)d⇒ d=14πε02Ze2K

Since the nuclear charge cannot be spread wider than the region the alpha never entered, d is an upper limit on the radius (Fig. 13.5). A handy number for such sums is e2/4πε0=1.44 MeV fm, where 1 fm=10−15 m (one femtometre, also called a fermi).

2/3All its kinetic energy has become electric potential energy at 29.5 fm, so it stops and turns back.

Fig. 13.5Alphas of a few MeV turn back far outside a gold nucleus; only much more energetic ones reach its surface and feel the nuclear force.NCERT p. 309

A large orange circle on the right is a gold nucleus drawn to scale, with a dashed line running left from its centre and a 10 femtometre scale bar below. Step 1: a small purple alpha particle at the far left moves towards the nucleus with K equal to 7.7 MeV. Step 2: the alpha particle has stopped at the point where it turns back, and a dimension arrow from it to the centre of the nucleus reads d equals 29.5 fm, well outside the nucleus. Step 3: a dashed alpha particle with K of about 26 MeV is shown touching the surface of the nucleus, labelled touching at 8.9 fm, where the nuclear pull starts.

Example

How close do Rutherford's alphas get?

Rutherford's best alpha source gave particles of kinetic energy 7.7 MeV. (a) Find the distance of closest approach of such an alpha particle to a gold nucleus (Z=79) in a head-on collision. (b) Taking nuclear radii from R=1.2A1/3 fm (Eq. 13.5 below), how much energy would an alpha particle need to just touch the gold nucleus (A=197)?

Try it yourself first. Then reveal the solution one step at a time.

With faster projectiles the story changes. Once the projectile reaches the edge of the nucleus, the short-range nuclear force joins in and the scattering stops following Rutherford's formula; the energy at which this deviation begins gives the nuclear size. Even better probes are fast electrons. An electron does not feel the nuclear force at all, so its scattering maps out how the nuclear charge (the protons) is spread. Alpha particles and neutrons, which do feel the nuclear force, map the nuclear matter instead, so the two methods give slightly different radii. Experiments of both kinds, on many elements, agree on a remarkably simple rule:

Nuclear radius
R=R0A1/3
(13.5)
where
R0
1.2 fm=1.2×10−15 m, the same for every nucleus
A
the mass number

The cube root is the clue. The volume of a sphere goes as R3, so the volume of a nucleus goes as R03A: it is simply proportional to the number of nucleons. Doubling the nucleons doubles the volume, but stretches the radius only by 21/3=1.26 (Fig. 13.6). Nucleons in a nucleus are packed like marbles in a jar or molecules in a drop of water: adding more makes a bigger drop of the same density. This liquid-drop picture of the nucleus will help us understand fission later.

2/3Their radii are 1.2 fm × 2, 3, 4 and 5: the cube roots of A.

Fig. 13.6Multiply the nucleons by 125/8 ≈ 16 and the radius only grows from 2.4 fm to 6.0 fm: volume is proportional to A, so every nucleus has the same density.

Four circles stand on a baseline, growing from left to right, labelled A equals 8, 27, 64 and 125. Step 1 shows only the circles and their mass numbers. Step 2 adds their radii, 2.4, 3.6, 4.8 and 6.0 femtometres, the cube roots 2, 3, 4 and 5 times 1.2 fm, with a note that the radius grows as the cube root of A and the formula R equals R nought A to the one-third. Step 3 fills every circle with the same pattern of tightly packed dots and states: same packing, same density.

DerivationDensity of nuclear matter

The mass of a nucleus with mass number A is very nearly A atomic mass units, m≈A mu with mu=1.66×10−27 kg. Its volume follows from Eq. 13.5.

ρ=A mu43πR03A=3 mu4πR03

The mass number cancels, so every nucleus has the same density. Putting in the numbers,

ρ=3×1.66×10−27 kg4π (1.2×10−15 m)3≈2.3×1017 kg m−3

Compare gold, one of the densest metals at 1.9×104 kg m−3: nuclear matter is ten trillion times denser. In a gold bar the nuclei fill less than one part in 1013 of the volume; everything else is the thin haze of electrons.

Example 13.1NCERT Example 13.1 · p. 310

How dense is the core of an aluminium atom?

An aluminium-27 atom has a mass of 26.98 u. (a) Find the radius of its nucleus. (b) Find the density of the nucleus. (c) Aluminium metal has a density of 2700 kg m−3. How many times denser is the nucleus?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
What is the radius of a copper-64 nucleus, in femtometres?
fm
Show answer

Answer 4.8 fm

R=1.2×641/3=1.2×4=4.8 fm.

13.4 Mass-energy and nuclear binding energy

13.4.1 Mass – Energy

Before 1905, physics had two separate bookkeeping rules: the total mass in a reaction stays the same, and so does the total energy. Einstein's special theory of relativity merged them. Mass, he showed, is itself a form of energy, and it can be turned into kinetic energy, light or heat, and back again. The exchange rate is fixed by the speed of light c in vacuum:

Mass–energy equivalence
E=mc2
(13.6)
where
E
energy equivalent of the mass
m
mass
c
speed of light in vacuum, 3.00×108 m s−1

Because c2 is about 9×1016 m2 s−2, a little mass is worth an enormous amount of energy.

Example 13.2NCERT Example 13.2 · p. 310

A speck of dust as fuel

Suppose all the mass of a speck of dust, 1.0 mg, could be turned into energy. (a) How much energy is that? (b) For how long could it run a 1.5 kW air conditioner?

Try it yourself first. Then reveal the solution one step at a time.

Nobody can turn dust wholly into energy, but nuclear reactions and the creation and annihilation of particles test Eq. 13.6 directly, and it has passed every test. The modern statement is a single law: in any isolated process the total energy is conserved, provided we count the energy mc2 of every mass. There are no longer separate laws for mass and energy, only one law of conservation of mass–energy. Nuclear physics gives the most striking evidence for it, and the rest of this chapter runs on it.

Quick check
1 of 2
How much energy, in joules, is equivalent to a mass of 2.0 g?
J
Show answer

Answer 180000000000000 J

E=mc2=(2.0×10−3 kg)(3.0×108 m s−1)2=1.8×1014 J.

13.4.2 Nuclear binding energy

A nucleus is made of protons and neutrons, so you might expect its mass to equal the masses of its parts added up. Weigh them and it does not. Take carbon-12, whose atom has a mass of exactly 12.000000 u by definition. It has 6 protons, 6 neutrons and 6 electrons. A neat trick handles the electrons: count the protons and electrons together as 6 hydrogen atoms, so that the electrons appear on both sides and cancel. The parts then weigh

6 mH+6 mn=6.046950 u+6.051990 u=12.098940 u

The assembled atom is lighter by 12.098940−12.000000=0.098940 u, almost a tenth of a nucleon's mass (Fig. 13.7). Every nucleus shows such a shortfall, called its mass defect (द्रव्यमान क्षति). For a nucleus of mass M with Z protons and A−Z neutrons,

Mass defect
ΔM=[ Zmp+(A−Z) mn ]−M
(13.7)

With atomic masses from tables, replace mp by mH and M by the atomic mass, as we did for carbon; the electron masses cancel. (The tiny binding energies of the electrons themselves, a few keV at most, are ignored.)

3/3The missing mass left as energy, ΔMc2=92.2 MeV, when the nucleus formed.

Fig. 13.7The separate parts of a carbon-12 atom are heavier than the atom itself; the missing mass is the binding energy divided by c².NCERT p. 311

A beam balance on a stand. The left pan holds twelve small balls, six marked H for hydrogen atoms and six marked n for neutrons, labelled 6 H atoms plus 6 n, 12.098 940 u. The right pan holds one compact cluster labelled one C-12 atom, 12.000 000 u. Step 1 shows the balance level, as if the two sides were equal. Step 2 tips the beam so the left pan drops, with the caption missing mass delta M equals 0.098 940 u. Step 3 keeps the tilt and adds a highlighted card: binding energy 92.2 MeV.

Where has the mass gone? Mass–energy equivalence answers this. When 6 free protons and 6 free neutrons come together to form a carbon nucleus, they give out energy, and that energy carries off mass ΔM. Run the process backwards: to pull the nucleus apart into separate nucleons you must supply the same energy, and the separated parts are then heavier by exactly ΔM. This energy is the binding energy (बंधन ऊर्जा) of the nucleus:

Binding energy
Eb=ΔM c2
(13.8)

Binding energy is a bookkeeping quantity: experiments rarely smash a nucleus into single nucleons, yet comparing binding energies tells us which nuclei are held more firmly and which reactions can release energy. To use Eq. 13.8 with masses in u, we need to know what one atomic mass unit is worth in energy.

Example 13.3NCERT Example 13.3 · p. 311

The energy in one atomic mass unit

(a) Find the energy equivalent of an electron mass (9.109×10−31 kg) and of 1 u, in joules and in MeV. Take c=2.998×108 m s−1 and 1 eV=1.602×10−19 J. (b) Use the result to find the binding energy of helium-4 (atomic mass 4.002603 u), in total and per nucleon.

Try it yourself first. Then reveal the solution one step at a time.

For carbon-12 the same conversion gives Eb=0.098940×931.5 MeV=92.2 MeV. Big nuclei naturally have big binding energies simply because they have more nucleons, so to compare nuclei fairly we divide by A. The binding energy per nucleon,

Binding energy per nucleon
Ebn=EbA
(13.9)

is each nucleon's share of the binding: the total cost of dismantling the nucleus, divided equally among its A members. Carbon-12 has 92.2/12=7.68 MeV per nucleon. Work this out for every stable nucleus, plot it against the mass number, and a famous curve appears (Fig. 13.8).

2/4From A≈30 to 170 every nucleon is bound by about 8 to 8.8 MeV; the peak is near A=56 to 62.

Fig. 13.8Binding energy per nucleon is nearly constant for middle-sized nuclei and lower at both ends, so energy is released both by splitting very heavy nuclei and by joining very light ones.NCERT Fig. 13.1 · p. 312

A graph with mass number A from 0 to 250 across and binding energy per nucleon from 0 to 9 MeV up. Blue dots for about 35 stable nuclei follow a curve: H-2 sits near 1 MeV, helium-4 jumps to about 7 MeV above its lighter neighbours, carbon-12 and oxygen-16 also sit slightly high, the curve flattens above A of about 30, peaks near 8.8 MeV at iron-56, and slides gently to about 7.6 MeV at uranium-238. Step 1 shows the points and the trend line. Step 2 draws a dashed box over the nearly flat stretch from A of 30 to 170, labelled nearly flat, about 8 to 8.8. Step 3 shades the heavy side and draws an arrow from A of about 236 to A of about 118 labelled fission, heavy splits. Step 4 shades the light side and draws an arrow from H-2 up to helium-4 labelled fusion, light nuclei join.

Read the curve from left to right. For the lightest nuclei Ebn is small (1.1 MeV for the deuteron) and rises steeply, with a few nuclei sitting noticeably above their neighbours: helium-4 (7.07 MeV), carbon-12 and oxygen-16. These favoured nuclei are evidence that nucleons fill shells inside the nucleus, much as electrons fill shells in an atom. Beyond about A=30 the curve flattens out between 8 and 8.8 MeV per nucleon all the way to A≈170, with its highest point around iron-56 and nickel-62 at about 8.8 MeV. For the heaviest nuclei it sags slowly, to 7.57 MeV for uranium-238.

Four conclusions follow, and they shape the rest of the chapter.

  1. The binding force is strongly attractive. Holding each nucleon with several MeV, against the electric repulsion of the protons, needs a force much stronger than any we have met so far.
  2. The force has a short range. If every nucleon attracted every other one, each new nucleon would add to the binding of all the rest, and Ebn would keep growing with A. Instead it stays flat, which means a nucleon interacts only with the few neighbours within reach.
  3. Splitting a heavy nucleus releases energy. Its nucleons, bound by about 7.6 MeV each, end up in middle-sized nuclei where they are bound by about 8.4 to 8.7 MeV. Every nucleon becomes more tightly bound, and the difference comes out as energy: this is fission (Section 13.7.1).
  4. Joining light nuclei releases energy. Two deuterons (1.1 MeV per nucleon) that combine into helium-4 (7.1 MeV per nucleon) end up far more tightly bound. This is fusion, the energy source of the Sun (Section 13.7.2).

The second conclusion deserves a closer look (Fig. 13.9). Suppose the pull reaches only about 2 fm, roughly as far as the nearest neighbours. Then a nucleon buried inside a nucleus is surrounded by the same small crowd, about a dozen nucleons in three dimensions, whether the nucleus has 40 nucleons or 200. Nucleons added on the far side are out of reach and change nothing for it. In a large nucleus most nucleons are buried in this way, so Ebn settles to a constant. Only the nucleons at the surface, with fewer neighbours, are bound less; in a small nucleus nearly all nucleons are at the surface, which is why Ebn is lower for light nuclei. This property of a short-range force, that each nucleon is held only by a limited number of neighbours however large the nucleus, is called saturation.

1/3The pull on a nucleon comes only from neighbours inside the dashed circle.

Fig. 13.9A nucleon feels only the neighbours within the short range of the nuclear force, so interior nucleons are equally well bound in every large nucleus, while surface nucleons are bound less.

On the left a small cluster of seven grey discs; on the right a large hexagonal cluster of thirty-seven. Step 1: in the large cluster the central nucleon is coloured purple, a dashed circle marks the range of the pull, about 2 fm, and its six nearest neighbours inside the circle are shaded. Step 2: a nucleon at the far edge of the cluster is outlined with a dashed border and labelled too far: no pull, and the heading says same neighbours, any A. Step 3: the dashed circle is drawn around a corner nucleon on the surface, which has only three neighbours inside it, the heading says surface nucleon: only 3, and the small cluster is shaded and labelled mostly surface.

Quick check
1 of 4
Oxygen-16 has an atomic mass of 15.994915 u. Using mH=1.007825 u, mn=1.008665 u and 1 u=931.5 MeV/c2, find its binding energy per nucleon in MeV.
MeV
Show answer

Answer 7.98 MeV

ΔM=8(1.007825)+8(1.008665)−15.994915=0.137005 u, so Eb=0.137005×931.5=127.6 MeV and Ebn=127.6/16=7.98 MeV.

13.5 Nuclear force

Electrons are held in atoms by the Coulomb force; the outer electrons, which decide chemistry, are bound by only a few eV. Nucleons are held with about 8 MeV each, roughly a million times more, so a different force must be at work. Its job is hard. Two protons 2 fm apart repel each other with a force of F=e2/(4πε0r2)≈58 N, the weight of a 6 kg mass, acting on a particle of mass 10−27 kg. Their electric potential energy at that distance is about 0.72 MeV. Gravity is no help: between the same two protons it is about 1036 times weaker than the electric push. The force that wins is called the nuclear force (नाभिकीय बल), or the strong force. Experiments carried out mostly between 1930 and 1950 established its main features.

  1. It is much stronger than the Coulomb force at nuclear distances, strong enough to bind protons that repel one another into a volume a few femtometres across. Gravity, in turn, is far weaker than both.
  2. It has a short range. Between two nucleons the force fades away once they are more than a few femtometres apart. This is the origin of saturation and of the flat stretch of the binding-energy curve.
  3. It is attractive at moderate distances but strongly repulsive at very short ones. The potential energy of two nucleons is lowest at a separation r0 of about 0.8 fm. Farther apart than r0, they attract; pushed closer than r0, they repel very strongly. This hard core stops nuclei from collapsing.
  4. It does not depend on charge. The nuclear pull between two protons, two neutrons, or a proton and a neutron is about the same.

3/4At r0≈0.8 fm the potential energy is lowest and the force is zero.

Fig. 13.10Schematic potential energy of two nucleons: no force far apart, attraction down the slope towards the minimum at about 0.8 fm, and a hard repulsive core closer in.NCERT Fig. 13.2 · p. 313

At the top, a fixed blue nucleon on the left and an orange nucleon that moves along a line, joined by a dotted drop line to a red dot on a graph below. The graph shows potential energy U in MeV against separation r in femtometres from 0 to 4: the blue curve is flat at zero far out, dips to a minimum of about minus 100 MeV at 0.8 fm, and shoots steeply upwards below about 0.6 fm. Step 1: the nucleons are 3.3 fm apart, the dot sits on the flat part, labelled far apart: no force. Step 2: at 1.6 fm the dot is on the slope going down to the minimum, an arrow pulls the orange nucleon towards the blue one, labelled attraction. Step 3: at 0.8 fm the dot is at the bottom of the well, labelled lowest energy, at 0.8 fm. Step 4: at 0.55 fm the dot is high on the steep wall, an arrow pushes the orange nucleon away, labelled strong repulsion.

The force on a nucleon is the negative slope of the potential-energy curve, F=−dU/dr. Where U rises as r grows (beyond r0), F is negative, meaning it points towards smaller r: attraction. Where U rises as r shrinks (inside r0), the force points outwards: repulsion. Unlike Coulomb's law or Newton's law of gravitation, the nuclear force has no simple formula, and the curve in Fig. 13.10 is only a sketch of its shape.

Quick check
1 of 3
Two nucleons are 1.5 fm apart, where their potential energy rises as they move farther apart. The nuclear force between them is
Show answer

Answer (A) attractive

F=−dU/dr. If U increases with r, dU/dr>0 and F<0: the force points towards smaller r, an attraction. The force is charge-independent, so it is the same for any pair of nucleons.

13.6 Radioactivity

In February 1896 Henri Becquerel was testing whether uranium salts, after soaking up sunlight, give off penetrating rays like the X-rays discovered a few months earlier. A run of cloudy days in Paris stopped his experiment, so he left the salts on a photographic plate wrapped in black paper inside a drawer. When he developed the plate anyway, it was strongly fogged. The uranium was emitting radiation all by itself, with no sunlight needed. Marie and Pierre Curie soon found two far more active elements, polonium and radium, and Marie Curie named the effect radioactivity (रेडियोऐक्टिवता).

Later experiments showed that radioactivity comes from the nucleus. An unstable nucleus changes into another nucleus, or into a lower-energy state of itself, and throws out radiation in the process. This is radioactive decay. Three kinds occur in nature, named with the first three letters of the Greek alphabet before anyone knew what they were:

  • Alpha (α) decay: the nucleus emits a helium-4 nucleus, X24X2224He, two protons and two neutrons bound together.
  • Beta (β) decay: the nucleus emits an electron (β−) or a positron (β+), a particle with the electron's mass but charge +e. A neutrino or antineutrino leaves too.
  • Gamma (γ) decay: an excited nucleus drops to a lower energy state and emits a photon of high energy, typically hundreds of keV or more.
Table 13.1How each decay changes a nucleus (with one example of each)
DecayWhat leavesChangeExample
αX24X2224He nucleusZ − 2, N − 2, A − 4X88226X2882226Ra→X86222X2862222Rn+X24X2224He
β−electron, antineutrinoZ + 1, N − 1X614X26214C→X714X27214N+eX−+νˉ
β+positron, neutrinoZ − 1, N + 1X1122X211222Na→X1022X210222Ne+eX++ν
γphotonnoneX2860X228260NiX∗→X2860X228260Ni+γ

In every decay, charge and the total number of nucleons are conserved: add up the top numbers and the bottom numbers on each side of the examples in Table 13.1 and they match. Inside the nucleus, β− decay turns a neutron into a proton, and β+ decay turns a proton into a neutron. On a nuclide chart each decay is a fixed step (Fig. 13.11).

2/5Alpha decay removes 2 protons and 2 neutrons: two steps down and two to the left.

Fig. 13.11Each kind of decay moves a nucleus by a fixed step on the N–Z chart; gamma decay leaves it in the same cell.

A grid of cells with N increasing to the right and Z upwards; the parent nucleus, labelled Z, N, sits in the centre cell. Step 1 shows only the parent. Step 2: an arrow goes two cells left and two down to a dashed daughter cell, and the side panel reads Z minus 2, N minus 2, A minus 4: alpha decay. Step 3: an arrow goes one cell left and one up, reading Z plus 1, N minus 1, A same: beta-minus decay. Step 4: an arrow goes one cell right and one down, reading Z minus 1, N plus 1, A same: beta-plus decay. Step 5: a looping arrow returns to the same cell, reading Z, N, A unchanged; energy drops: gamma decay.

Why are some nuclei unstable at all? Light stable nuclei have about as many neutrons as protons (N/Z≈1). Heavier ones need extra neutrons: their many protons repel each other over the whole nucleus, while the nuclear attraction reaches only nearest neighbours, and extra neutrons add attraction without adding repulsion. By lead-208 the ratio is 126/82≈1.5. A nucleus with too many neutrons for its size tends to undergo β− decay, one with too many protons β+ decay, and very heavy nuclei often shed an alpha particle. Stability is the exception: of the more than 3000 nuclides known, fewer than one in ten is stable. The rest are made in laboratories by bombarding stable nuclei with protons, neutrons, deuterons or alpha particles, or are seen in the debris of stars.

Quick check
1 of 3
Uranium-238 (X92238X2922238U) undergoes alpha decay. The daughter nucleus is
Show answer

Answer (A) X90234X2902234Th

An alpha particle takes away 2 protons and 2 neutrons: Z=92−2=90 (thorium), A=238−4=234. Keeping A=238 forgets that the alpha particle carries 4 nucleons.

13.7 Nuclear energy

The binding-energy curve is also a map of where energy can be found. Its middle is a long plateau at about 8 MeV per nucleon, and both ends lie lower. A system that is more tightly bound has less mass, so whenever nucleons rearrange from loosely bound nuclei into more tightly bound ones, mass disappears and energy comes out. There are two ways to climb towards the plateau: split a very heavy nucleus into two middle-sized ones (fission), or join two very light nuclei into a heavier one (fusion).

The energy given out in a nuclear process is its Q-value. By conservation of mass–energy it can be found either from the kinetic energies or from the masses:

Q-value
Q=Kfinal−Kinitial=(∑minitial−∑mfinal)c2

A positive Q means energy is released (the process is exothermic); a negative Q means energy must be supplied (endothermic). Since protons and neutrons are only regrouped, Q also equals the total binding energy after minus the total binding energy before.

Compare this with chemistry. Coal, petrol and cooking gas release energy by rearranging electrons between atoms, a few eV per atom: burning one carbon atom into carbon dioxide gives about 4.1 eV. A single uranium fission gives about 200 MeV, roughly fifty million times more. Per kilogram, the complete fission of uranium-235 would give about 8×1013 J, while a kilogram of good coal gives about 3×107 J. Kilogram for kilogram, nuclear fuel is a few million times richer than chemical fuel. Explore the curve and both kinds of reaction in the lab below.

InteractiveBinding-energy curve lab

A graph of binding energy per nucleon against mass number for about 35 stable nuclides, with three modes. In Read mode a slider picks a nuclide and cards show its binding energy per nucleon, its total binding energy and its mass number. In Split mode a slider sets the mass number of the lighter fragment from a uranium-236 nucleus and a menu sets two or three free neutrons; markers and dashed lines show the parent and the two fragments on the curve, and cards give the total binding energy before and after and the energy released, about 160 to 205 MeV. In Join mode a menu picks one of seven fusion reactions, such as deuterium plus tritium, and the cards give the binding energies before and after and the Q-value.

Fig. 13.12Read binding energies off the curve, split a uranium-236 nucleus into fragments of your choice, or pick a fusion reaction, and see how much energy is released.

13.7.1 Fission

Natural radioactivity is slow and cannot be switched on or off. Far more control comes from nuclear reactions, in which we fire particles such as protons, deuterons, alpha particles or neutrons at nuclei. Neutrons make the best bullets: being uncharged, they feel no Coulomb repulsion and can reach a nucleus even when moving slowly. In 1938 Otto Hahn and Fritz Strassmann bombarded uranium with neutrons and, to their surprise, found barium among the products, an element about half as heavy. Lise Meitner and Otto Frisch explained it within weeks: the uranium nucleus had split in two. They called this nuclear fission (नाभिकीय विखंडन).

When uranium-235 absorbs a slow neutron, it becomes uranium-236 in a highly excited state, written X92236X2922236UX∗. This nucleus wobbles like a struck water drop and, in most cases, splits into two medium-sized fragments and a few neutrons (Fig. 13.13):

Neutron capture
X01X2021n+X92235X2922235U→X92236X2922236UX∗

The split is not the same every time. Three of the many ways the excited nucleus can break up are

X236X22236UX∗⟶X141X22141Ba+X92X2292Kr+3 n
(13.10)
X236X22236UX∗⟶X139X22139Xe+X95X2295Sr+2 n
(13.11)
X236X22236UX∗⟶X140X22140Cs+X93X2293Rb+3 n
(13.12)

In each case the charges balance too: barium, krypton, xenon, strontium, caesium and rubidium have Z=56,36,54,38,55 and 37, and each pair adds up to uranium's 92.

4/4Two fragments fly apart with 2 or 3 neutrons; about 200 MeV is released.

Fig. 13.13In the liquid-drop picture, the excited compound nucleus stretches until electric repulsion between its two halves wins over the nuclear attraction across the neck.

Step 1: a small blue ball marked n, labelled slow, moves towards a large orange circle labelled U-235. Step 2: the neutron has been absorbed and the drop is stretched sideways into an oval, labelled U-236 star: excited, wobbling. Step 3: the drop has a narrow waist in the middle like a peanut, labelled a neck forms. Step 4: two unequal circles, each marked fragment, have flown apart to the left and right, and three small neutrons fly out, labelled 2 fragments plus 2 or 3 neutrons plus about 200 MeV.

The liquid-drop picture explains why the split happens. As the excited drop stretches, the nuclear attraction, acting only between neighbours, holds the surface together less well, while the electric repulsion between its two ends, acting across the whole nucleus, stays strong. Past a certain stretch, repulsion wins. The fragments are born with roughly the neutron-to-proton ratio of uranium, about 1.55, which is too neutron-rich for medium nuclei. So they are radioactive and reach stability through a chain of β− decays, which is why spent reactor fuel is so dangerous.

Example

Estimating the energy of one fission

Use the binding-energy curve to estimate the energy released when X236X22236UX∗ splits into fragments with A=95 and A=139 and two free neutrons. Read Ebn≈7.59 MeV for A=236, 8.66 MeV for A=95 and 8.38 MeV for A=139.

Try it yourself first. Then reveal the solution one step at a time.

About 170 MeV of it is carried off at first by the two fragments, flung apart by their mutual electric repulsion. The rest goes to the neutrons, to gamma rays, and to the later beta decays of the fragments, including antineutrinos that escape. The fragments are stopped within a fraction of a millimetre in the surrounding fuel, so their energy becomes heat. That heat is the energy source of every nuclear power station: it boils water, and the steam turns turbines exactly as in a coal plant. Because each fission also releases 2 or 3 neutrons, one fission can trigger others. In a reactor this chain is held steady at one new fission per fission; in a nuclear weapon it is allowed to run away, releasing enormous energy in a fraction of a second.

Quick check
1 of 3
Why are neutrons, rather than protons, used to trigger fission in uranium?
Show answer

Answer (B) Neutrons feel no Coulomb repulsion from the nucleus

A proton approaching a uranium nucleus (Z=92) must climb a Coulomb barrier of more than 10 MeV; an uncharged neutron, even a slow one, simply walks in. Neutrons are actually slightly heavier than protons.

13.7.2 Nuclear fusion – energy generation in stars

On the left of the binding-energy curve, energy comes from joining nuclei. When two light nuclei combine into a heavier, more tightly bound one, the process is nuclear fusion (नाभिकीय संलयन). Some important fusion reactions, with Q-values worked out from atomic masses, are collected in Table 13.2.

Table 13.2Some fusion reactions and their Q-valuesNCERT Eq. 13.13 · p. 315
ReactionQ (MeV)Where it matters
X1X221H+X1X221H→X2X222H+eX++ν0.42first, slowest step in the Sun
X2X222H+X2X222H→X3X223He+n3.27deuterium fusion (one branch)
X2X222H+X2X222H→X3X223H+X1X221H4.03deuterium fusion (other branch)
X2X222H+X3X223H→X4X224He+n17.59fuel of experimental fusion reactors

Fusion has a catch. For the short-range nuclear force to grab the two nuclei, they must come within a few femtometres of each other, but both are positively charged and repel. They must therefore approach with enough kinetic energy to climb over the Coulomb barrier (Fig. 13.14). The height of the barrier grows with the charges of the two nuclei, so it is lowest, a few hundred keV, for the hydrogen isotopes.

1/4Outside contact, the electric repulsion makes a hill; inside, the nuclear force makes a deep well.

Fig. 13.14Two nuclei must get past their electric repulsion before the nuclear force can bind them; in stars a few unusually fast nuclei manage it, helped by quantum tunnelling.NCERT Points to ponder 5 · p. 316

A blue curve of potential energy against separation r. From the right the curve rises gently as the nuclei approach, a Coulomb hill whose top, at the contact distance, is labelled about 0.44 MeV for D–T; just inside, it drops vertically into a deep nuclear well. Step 1 shows only the curve. Step 2: a low dashed red energy line comes in from the right and stops far out on the hill, with arrows showing the nucleus turning back, labelled too slow: turns back. Step 3: a dashed energy line above the top of the hill runs all the way in, labelled fast enough: over the top. Step 4: a medium energy line reaches the hill and a dotted arrow passes straight through it into the well, labelled rare fast one tunnels.

Example

How hot must fusion fuel be?

A deuteron and a triton (the nucleus of tritium) fuse once they touch. Treat them as spheres with R=1.2A1/3 fm. (a) Find the height of the Coulomb barrier. (b) At what temperature would the average kinetic energy of a nucleus, 32kBT, equal this barrier? (kB=1.38×10−23 J K−1.)

Try it yourself first. Then reveal the solution one step at a time.

Fusion produced by heating matter until its nuclei move fast enough to overcome the barrier is called thermonuclear fusion. It powers every star. Yet the centre of the Sun is at about 1.5×107 K, where the average kinetic energy of a proton, 32kBT, is only about 2 keV. Two things rescue the Sun. Speeds in a hot gas are spread out, so a tiny fraction of protons move many times faster than average; and quantum mechanics lets such a proton tunnel through the top of the barrier without climbing over it. Each proton in the core waits on average billions of years to fuse, and that slowness is why the Sun shines steadily for so long.

The Sun turns hydrogen into helium through the proton–proton (p–p) chain (Fig. 13.15, Table 13.3):

Table 13.3The steps of the p–p chainNCERT Eq. 13.14 · p. 316
StepReactionQ (MeV)
(i)X1X221H+X1X221H→X2X222H+eX++ν0.42
(ii)eX++eX−→2 γ1.02
(iii)X2X222H+X1X221H→X3X223He+γ5.49
(iv)X3X223He+X3X223He→X4X224He+2 X1X221H12.86

Here X1X221H stands for a proton and X2X222H for a deuteron. In step (ii), each positron made in step (i) meets one of the plasma's free electrons and the pair annihilates.

Reaction (iv) consumes two helium-3 nuclei, so the first three reactions have to run twice for every once it runs. Doubling (i)–(iii) and adding (iv), the two protons handed back in (iv) cancel two of the six that went in, and the books balance as

Net result of the p–p chain
4 X1X221H+2 eX−⟶X4X224He+2ν+6γ+26.7 MeV
(13.15)

Adding two more electrons to each side turns this into a statement about whole atoms: four hydrogen atoms become one helium atom and release 26.7 MeV. Check it with atomic masses: 4(1.007825)−4.002603=0.028697 u, and 0.028697×931.5 MeV=26.7 MeV.

4/4Steps (i)–(iii) happen twice, and each positron annihilates with an electron (step ii).

Fig. 13.15In the p–p chain four protons end up as one helium-4 nucleus; counting each step and the two positron annihilations gives 26.7 MeV.

Balls marked plus are protons and balls marked n are neutrons. Step 1, top row: two single protons join to make a deuteron, one proton and one neutron, plus a positron and a neutrino, 0.42 MeV. Step 2 adds a second row: a deuteron and a proton make helium-3, two protons and a neutron, plus a gamma ray, 5.49 MeV. Step 3 adds a third row and dims the first two: two helium-3 nuclei make helium-4, two protons and two neutrons, and release two protons, 12.86 MeV. Step 4 marks the first two rows times two and adds a highlighted card with the tally: two times the sum of 0.42, 1.02 and 5.49, plus 12.86, equals 26.7 MeV.

Example

How fast does the Sun use its fuel?

The Sun radiates 3.8×1026 W. (a) How much mass does it lose each second? (b) How much hydrogen does it turn into helium each second, if each 26.7 MeV comes from four hydrogen atoms (atomic mass 1.0078 u)? Ignore the small share of energy carried off by neutrinos.

Try it yourself first. Then reveal the solution one step at a time.

The Sun is about 4.6 billion years old and has enough hydrogen in its core to keep going for about another 5 billion. When the core hydrogen runs low, fusion there slows, and the core contracts under gravity and heats up. At about 108 K helium nuclei themselves begin to fuse: three helium-4 nuclei make one carbon-12, releasing 7.3 MeV. Meanwhile the outer layers swell enormously and the Sun becomes a red giant. In stars much heavier than the Sun the core gets hotter still and builds oxygen, neon, silicon and finally iron. There the chain stops, because iron sits at the top of the binding-energy curve: making anything heavier by fusion absorbs energy. Elements beyond iron, such as the gold in jewellery and the uranium in reactors, were built by neutron capture in exploding stars and in collisions of neutron stars.

Quick check
1 of 3
Why does fusion need temperatures of millions of kelvin, while fission can be started with slow neutrons?
Show answer

Answer (D) Two positive nuclei must overcome their electric repulsion to get close

Both fusing partners are positively charged, so they must approach with enough kinetic energy to get past the Coulomb barrier; heat supplies that energy. A neutron triggering fission has no charge and faces no barrier.

13.7.3 Controlled thermonuclear fusion

Could we build a star on Earth? A fusion power plant would burn deuterium, which is plentiful in sea water, with tritium, which can be bred from lithium, and would leave no long-lived fission fragments. The fuel must be heated to around 108 K, several times hotter than the Sun's core, because a reactor cannot wait billions of years for rare tunnelling events. At such temperatures the atoms are stripped of their electrons, and the fuel becomes a plasma, a gas of ions and free electrons. The challenge is to hold it: any solid wall would cool the plasma at once and itself be destroyed. In a tokamak, strong magnetic fields shaped like a doughnut keep the charged particles circling away from the walls (recall from Chapter 4 of Part I how a magnetic field bends moving charges). Another approach crushes tiny fuel pellets with powerful laser beams.

Example 13.4NCERT Example 13.4 · p. 317

Three claims to check

Test each claim. (a) Eq. 13.10 cannot be balanced, because uranium goes in and barium and krypton come out. Compare it with a chemical equation such as CHX4+2 OX2→COX2+2 HX2O. (b) If every proton and neutron survives a reaction, no mass can disappear. (c) Mass turns into energy only in nuclear reactions, never when fuel burns. Burning 12 g of carbon releases 3.94×105 J; compare the fraction of mass that becomes energy with that in one fission of uranium-235 (200 MeV, A=236).

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
A fusion reactor plasma is held by magnetic fields rather than by material walls mainly because
Show answer

Answer (A) any wall in contact would cool the plasma and be damaged

Contact with a wall would cool the plasma below fusion temperature at once (and the wall would erode). Magnetic fields act on the charged ions and electrons without touching them. The nuclei in a plasma are charged, not neutral.
Recap

Nuclei: key points and summary

  1. The nucleus is about 104 times smaller than the atom in radius but holds more than 99.9 % of its mass (Section 13.1).
  2. Atomic masses are measured in atomic mass units: 1 u is one-twelfth of the mass of a carbon-12 atom, 1.6605×10−27 kg (Section 13.2).
  3. A nucleus contains Z protons and N neutrons; the mass number is A=Z+N and the nuclide is written XZAX2Z2AX. The neutron, found by Chadwick in 1932, is slightly heavier than the proton (Discovery of neutron).
  4. Isotopes share Z, isobars share A, isotones share N. An element's atomic mass is the abundance-weighted average of its isotope masses (Section 13.2).
  5. Scattering experiments give R=R0A1/3 with R0=1.2 fm, so all nuclei have the same density, about 2.3×1017 kg m−3 (Section 13.3).
  6. Mass is a form of energy, E=mc2; total energy, including mc2, is conserved (Section 13.4.1).
  7. A nucleus is lighter than its separate nucleons by the mass defect ΔM=[Zmp+(A−Z)mn]−M; its binding energy is Eb=ΔMc2, with 1 u≡931.5 MeV (Section 13.4.2).
  8. Binding energy per nucleon is about 8 MeV for 30<A<170, peaks near 8.8 MeV around iron and is lower for light and very heavy nuclei (Section 13.4.2).
  9. The nuclear force is strong, short-ranged (a few fm), repulsive below about 0.8 fm and independent of charge; its short range explains saturation (Section 13.5).
  10. Radioactive nuclei emit alpha particles, electrons or positrons, or gamma rays; charge and nucleon number are conserved in every decay (Section 13.6).
  11. The Q-value of a process is (∑minitial−∑mfinal)c2. Nuclear energies are MeV per event against eV for chemistry, a million-fold difference (Section 13.7).
  12. Fission of a heavy nucleus such as uranium-235 releases about 200 MeV; fusion of light nuclei powers the stars, with the p–p chain turning four hydrogen atoms into helium and 26.7 MeV, and needs very high temperatures to beat the Coulomb barrier (Section 13.7.1, Section 13.7.2, Section 13.7.3).
Table 13.4Quantities in this chapterNCERT p. 319
QuantitySymbolUnitTypical size
Atomic mass unitu1 u=1.6605×10−27 kgproton 1.0073 u
Nuclear radiusR=R0A1/3fm (10−15 m)1.2 fm (A = 1) to 7.4 fm (A = 238)
Mass defectΔMu0.030 u (helium-4) to 1.93 u (uranium-238)
Binding energyEbMeV2.2 MeV (deuteron) to 1800 MeV (uranium-238)
Binding energy per nucleonEbnMeV1.1 MeV (deuteron) to 8.8 MeV (iron, nickel)
Q-valueQMeV0.42 MeV (p + p) to about 200 MeV (fission)
Decay constantλs−1chance of decay per second
Half-lifeT1/2s5730 years for carbon-14
Mean lifeτsabout 880 s for a free neutron
ActivityRBq (decays per second)grows with the number of nuclei and with λ
Flash cards · 10

Formula revision

1 / 10
Practice · 16 questions

Practice questions with answers (NCERT “Exercises”)

  1. NumericalEasyNCERT Exercises 13.1

    Neon-20 has an atomic mass of 19.992440 u. Find its binding energy in MeV. Where needed, use: mH=1.007825 u, mn=1.008665 u, 1 u=931.5 MeV/c2, 1 MeV=1.602×10−13 J, NA=6.022×1023 mol−1, e2/4πε0=1.44 MeV fm, R0=1.2 fm, 1 year =3.156×107 s.

    MeV
    Show solution

    Neon-20 has 10 protons and 10 neutrons. Using atomic masses, the electrons cancel:

    ΔM=10(1.007825)+10(1.008665)−19.992440=0.172460 u
    Eb=0.172460×931.5 MeV=160.6 MeV

    That is 8.03 MeV per nucleon.

  2. NumericalMediumNCERT Exercises 13.2

    Calcium-40 (atomic mass 39.962591 u) and lead-208 (atomic mass 207.976652 u) are both very stable nuclei.

    1. Find the binding energy of calcium-40 in MeV.

      MeV
      Show solution
      ΔM=20(1.007825)+20(1.008665)−39.962591=0.367209 u
      Eb=0.367209×931.5=342.1 MeV
    2. Find the binding energy of lead-208 in MeV.

      MeV
      Show solution
      ΔM=82(1.007825)+126(1.008665)−207.976652=1.756788 u
      Eb=1.756788×931.5=1636 MeV
    3. Which nucleus holds its nucleons more tightly?

      Show solution

      Per nucleon: calcium-40, 342.1/40=8.55 MeV; lead-208, 1636/208=7.87 MeV. Calcium-40 is more tightly bound, although lead-208 has the larger total binding energy.

  3. NumericalMediumNCERT Exercises 13.3

    A 5.0 g strip of aluminium foil is pure aluminium-27 (atomic mass 26.981539 u). How much energy would be needed to separate every proton and neutron in it from every other? Compare it with the energy of a 1-kiloton explosion, 4.2×1012 J.

    J
    Show solution

    Binding energy of one aluminium-27 nucleus (13 protons, 14 neutrons):

    ΔM=13(1.007825)+14(1.008665)−26.981539=0.241496 u,Eb=0.241496×931.5=225.0 MeV

    Number of atoms in 5.0 g:

    N=5.0 g26.98 g mol−1×6.022×1023 mol−1=1.116×1023
    E=NEb=1.116×1023×225.0 MeV=2.51×1025 MeV=4.0×1012 J

    That is about the energy of a 1-kiloton explosion, needed just to take apart the nuclei in a few grams of foil.

  4. NumericalEasyNCERT Exercises 13.4

    Estimate the ratio of the radius of a uranium-238 nucleus to that of a zinc-64 nucleus. What is the ratio of their volumes?

    Show solution
    RURZn=(23864)1/3=(3.72)1/3=1.55

    The volumes are in the ratio of the mass numbers, 238/64=3.72, since volume is proportional to A.

  5. NumericalMediumNCERT Exercises 13.5

    Two historic reactions are listed below. Using Q=(total mass before−total mass after)c2 and the atomic masses X1X221H 1.007825 u, X4X224He 4.002603 u, X7X227Li 7.016003 u, X14X2214N 14.003074 u and X17X2217O 16.999132 u, find Q for each historic reaction below and say whether it releases or absorbs energy.

    1. X37X2327Li+X11X2121H→X24X2224He+X24X2224He (the first nucleus split with an accelerator, Cockcroft and Walton, 1932). Give Q in MeV.

      MeV
      Show solution
      Δm=7.016003+1.007825−2(4.002603)=0.018622 u
      Q=0.018622×931.5=+17.35 MeV

      Q>0: exothermic. Electron masses balance because the atomic numbers add up the same on both sides (3+1=2+2).

    2. X714X27214N+X24X2224He→X817X28217O+X11X2121H (the first artificial transmutation, Rutherford, 1919). Give Q in MeV.

      MeV
      Show solution
      Δm=14.003074+4.002603−16.999132−1.007825=−0.001280 u
      Q=−0.001280×931.5=−1.19 MeV

      Q<0: endothermic. The alpha particle must bring at least this much kinetic energy (in fact more, because the products must also carry away momentum).

  6. NumericalMediumNCERT Exercises 13.6

    Could a calcium-40 nucleus (atomic mass 39.962591 u) split into two neon-20 nuclei (atomic mass 19.992440 u each) and release energy? Decide by finding Q.

    MeV
    Show solution
    Q=[39.962591−2(19.992440)]×931.5 MeV=(−0.022289)(931.5) MeV=−20.8 MeV

    Q is negative: the two neon nuclei together are heavier than the calcium nucleus. The split would need about 21 MeV of energy put in, so it cannot happen on its own. This fits the binding-energy curve: calcium-40 (8.55 MeV per nucleon) is more tightly bound than neon-20 (8.03 MeV per nucleon), and only nuclei well beyond the peak gain by splitting.

  7. NumericalMediumNCERT Exercises 13.7

    India's planned thorium reactors will burn uranium-233, which releases about 200 MeV per fission. (a) How much energy is released if every nucleus in 1.0 g of uranium-233 undergoes fission? Give the answer in MeV and in joules. (b) How many tonnes of coal (3.0×107 J per kg) would release the same energy?

    J
    Show solution
    N=1.0 g233 g mol−1×6.022×1023 mol−1=2.58×1021
    E=2.58×1021×200 MeV=5.17×1023 MeV=8.3×1010 J
    mcoal=8.3×1010 J3.0×107 J kg−1=2.8×103 kg

    One gram of uranium, a bead about 5 mm across, matches almost 3 tonnes of coal.

  8. NumericalMediumNCERT Exercises 13.8

    A fusion fuel pellet holds 1.0 g of a deuterium–tritium mixture with equal numbers of D and T atoms (atomic masses 2.014102 u and 3.016049 u). Each D–T fusion releases 17.6 MeV. If all of it fused, for how long could the energy run a 2.0 kW water heater?

    years
    Show solution

    Each fusion uses one D and one T, a pair of mass 2.014102+3.016049=5.030151 u. Number of pairs in 1.0 g:

    n=1.0 g5.030 g mol−1×6.022×1023=1.197×1023
    E=1.197×1023×17.6×1.602×10−13 J=3.38×1011 J
    t=3.38×1011 J2000 W=1.69×108 s≈5.3 years
  9. NumericalMediumNCERT Exercises 13.9

    Treat two alpha particles as hard spheres of radius 1.9 fm. Find the height of the Coulomb barrier they must overcome to touch head-on, in MeV. Use 1/4πε0=9×109 N m2 C−2 and e=1.6×10−19 C.

    MeV
    Show solution

    Each alpha particle has charge 2e; touching, their centres are 2×1.9=3.8 fm apart. The barrier is the electric potential energy at that distance:

    U=14πε0(2e)(2e)r=(9×109)(4)(1.6×10−19)23.8×10−15 J=2.43×10−13 J
    U=2.43×10−131.6×10−13 MeV=1.5 MeV

    This is several times higher than for two hydrogen nuclei, which is why helium burns only in the much hotter cores of ageing stars.

  10. ProofMediumNCERT Exercises 13.10

    Starting from R=R0A1/3:

    1. Show that the mass density of nuclear matter does not depend on A.

      Show solution

      The mass of a nucleus is about Amu and its volume is 43πR03A, so

      ρ=A mu43πR03A=3mu4πR03

      A cancels: every nucleus has the same density, about 2.3×1017 kg m−3.

    2. How many nucleons are there per cubic femtometre of nuclear matter?

      per fm³
      Show solution
      n=A43πR03A=34π(1.2 fm)3=0.138 fm−3
    3. Estimate the typical distance between neighbouring nucleons and compare it with the range of the nuclear force.

      Show solution

      Each nucleon has about 1/n=7.2 fm3 to itself, a cube of side n−1/3=1.9 fm. Neighbours sit about 2 fm apart, just inside the range of the nuclear attraction and outside its repulsive core (0.8 fm), as saturation requires.

  11. Quick answerEasy

    Which of these pairs are isobars?

    Show answer

    Answer (C) X614X26214C and X714X27214N

    Isobars share the mass number: carbon-14 and nitrogen-14 both have A=14. The potassium pair are isotopes, and carbon-13 and nitrogen-14 are isotones (N=7).
  12. Quick answerEasy

    What energy, in MeV, is equivalent to a mass of 0.0100 u?

    MeV
    Show answer

    Answer 9.315 MeV

    0.0100×931.5 MeV=9.32 MeV.
  13. Quick answerMedium

    Put these lengths in order from the smallest to the largest.

    1. closest approach of a 7.7 MeV alpha to gold
    2. radius of a gold atom
    3. radius of a gold nucleus
    4. radius of a proton (about 1 fm)
    Show answer
    Answer
    1. radius of a proton (about 1 fm)
    2. radius of a gold nucleus
    3. closest approach of a 7.7 MeV alpha to gold
    4. radius of a gold atom
    About 0.9 fm, 7.0 fm, 30 fm and 1.4×105 fm (0.14 nm).
  14. Quick answerEasy

    Match each decay with what it does to the nucleus.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • α decay : Z falls by 2, A by 4
    • β− decay : Z rises by 1, A unchanged
    • β+ decay : Z falls by 1, A unchanged
    • γ decay : Z and A unchanged
    An alpha particle carries 2 protons and 2 neutrons; beta decay changes one nucleon into the other kind; gamma decay only lowers the energy.
  15. Quick answerEasy

    One atomic mass unit is equivalent to MeV of energy (to four significant figures).

    Show answer

    Answer 931.5

    (1 u)c2=1.4924×10−10 J=931.5 MeV.
  16. Quick answerEasy

    True or false: in a fusion reaction that releases energy, the products have less total mass than the reactants.

    Show answer

    Answer True

    Released energy Q comes from mass: ∑minitial−∑mfinal=Q/c2>0.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Exercises 13.1Practice Q1
NCERT Exercises 13.2Practice Q2
NCERT Exercises 13.3Practice Q3
NCERT Exercises 13.4Practice Q4
NCERT Exercises 13.5Practice Q5
NCERT Exercises 13.6Practice Q6
NCERT Exercises 13.7Practice Q7
NCERT Exercises 13.8Practice Q8
NCERT Exercises 13.9Practice Q9
NCERT Exercises 13.10Practice Q10

Quick check: MCQs on Nuclei

  1. 13.1 Introduction2 questions
  2. Discovery of neutron4 questions
  3. 13.3 Size of the nucleus3 questions
  4. 13.4.1 Mass – Energy2 questions
  5. 13.4.2 Nuclear binding energy4 questions
  6. 13.5 Nuclear force3 questions
  7. 13.6 Radioactivity3 questions
  8. 13.7.1 Fission3 questions
  9. 13.7.2 Nuclear fusion – energy generation in stars3 questions
  10. 13.7.3 Controlled thermonuclear fusion2 questions
Glossary · 15

Key terms (English–Hindi)

atomic mass unitपरमाणु द्रव्यमान मात्रकparamanu dravyamaan maatrak
One-twelfth of the mass of a carbon-12 atom: 1 u=1.6605×10−27 kg, equivalent to 931.5 MeV.First seen in the chapter →
binding energyबंधन ऊर्जाbandhan oorja
The energy needed to separate a nucleus into free nucleons, equal to the energy released when it forms: Eb=ΔMc2.First seen in the chapter →
binding energy per nucleonप्रति न्यूक्लिऑन बंधन ऊर्जाprati nyookliyon bandhan oorja
Eb/A, the average binding of one nucleon; about 8 MeV for most nuclei and highest (8.8 MeV) near iron.First seen in the chapter →
isobarसमभारिकsambhaarik
Nuclides with the same mass number A but different Z.First seen in the chapter →
isotoneसमन्यूट्रॉनिकsamanyootronik
Nuclides with the same neutron number N but different Z.First seen in the chapter →
isotopeसमस्थानिकsamasthanik
Nuclides of one element that differ only in their neutron count: equal Z, unequal N and A.First seen in the chapter →
mass defectद्रव्यमान क्षतिdravyamaan kshati
How much lighter a nucleus is than its separate protons and neutrons: ΔM=[Zmp+(A−Z)mn]−M.First seen in the chapter →
mass numberद्रव्यमान संख्याdravyamaan sankhya
The total number of nucleons in a nucleus, A=Z+N.First seen in the chapter →
nuclear fissionनाभिकीय विखंडनnaabhikiya vikhandan
The splitting of a heavy nucleus, such as uranium-235 after absorbing a neutron, into two medium nuclei and a few neutrons, releasing about 200 MeV.First seen in the chapter →
nuclear forceनाभिकीय बलnaabhikiya bal
The strong, short-range, charge-independent force that binds nucleons; it vanishes beyond a few femtometres and repels below about 0.8 fm.First seen in the chapter →
nuclear fusionनाभिकीय संलयनnaabhikiya sanlayan
The joining of light nuclei into a heavier, more tightly bound one, releasing energy; it powers the stars.First seen in the chapter →
nucleonन्यूक्लिऑनnyookliyon
A proton or a neutron.First seen in the chapter →
nucleusनाभिकnaabhik
The tiny, dense, positively charged centre of an atom, made of protons and neutrons; it holds more than 99.9 % of the atom's mass.First seen in the chapter →
Q-valueQ-मानQ-maan
The energy released in a nuclear process: (∑minitial−∑mfinal)c2, equal to the gain in kinetic energy.First seen in the chapter →
radioactivityरेडियोऐक्टिवताrediyoaiktivata
The spontaneous decay of unstable nuclei with the emission of alpha particles, electrons or positrons, or gamma rays.First seen in the chapter →