Physics by Lamhi: Not Your Boring Physics

13.1 Nuclear Size & Density

Scattering experiments show that a nucleus is, to a good approximation, a sphere whose radius grows with the number of nucleons it contains:

Nuclear radius

R = R_0 A^{1/3}

where $A$ is the mass number and $R_0 \approx 1.2\text{ fm}$ (1 fm = 10⁻¹⁵ m) is roughly the same for every nucleus. Since volume scales as $R^3 \propto A$ , and mass also scales roughly as $A$ (every nucleon has about the same mass), the nuclear density comes out independent of A, every nucleus, light or heavy, is packed to roughly the same enormous density, about $2.3\times10^{17}\text{ kg/m}^3$ .

Worked example

Nuclear radius from mass number

Estimate the radius of an aluminium nucleus, $^{27}_{13}\text{Al}$ ( $A=27$ ).

R = R_0A^{1/3} = 1.2\times27^{1/3} = 1.2\times3 = \textcolor{#e08a1e}{3.6\text{ fm}}

13.1.1 The nuclear force: what actually holds a nucleus together

A nucleus packs many protons into a space just a few femtometres across, all mutually repelling each other through the Coulomb force. Something stronger must be holding the whole thing together, a completely separate interaction called the nuclear force, with properties unlike gravity or electromagnetism:

Short-ranged. The nuclear force is powerfully attractive out to about 2–3 fm, roughly a nucleon’s own size, but falls to essentially zero beyond about 10 fm. Gravity and the Coulomb force, by contrast, reach out to infinity (just getting weaker, never vanishing). This short range is exactly why alpha particles need to get extremely close to a nucleus before the nuclear force can overpower Coulomb repulsion and bind, as §13.1’s companion scattering experiments (see the Atoms chapter’s closest-approach calculation) quietly depend on.

Stronger than Coulomb repulsion at nuclear separations. At the 1–2 fm separations typical inside a nucleus, the nuclear force between two nucleons is roughly a hundred times stronger than the Coulomb repulsion between two protons at the same distance. That is precisely why a nucleus with many mutually repelling protons does not fly apart: at short range, the nuclear attraction simply wins.

Charge-independent. The nuclear force between two protons, two neutrons, or a proton and a neutron is the same strength (once the extra Coulomb repulsion between two protons is separately accounted for). The force itself does not care which of the two particles is charged, which is why a single $R=R_0A^{1/3}$ formula works for every nucleus, proton count and neutron count play symmetric roles in holding it together.

Saturating. A given nucleon only feels the nuclear force from its immediate neighbours, not from every other nucleon in the nucleus (unlike gravity or Coulomb forces, which add up over every pair). This is exactly why the binding energy per nucleon (§13.3) levels off rather than keeps climbing as $A$ grows, each nucleon only has so many neighbours to bind with, however large the nucleus gets.

13.2 Mass Defect & Binding Energy

The mass of a nucleus is always slightly less than the sum of the masses of its separate, free protons and neutrons. That missing mass is the mass defect:

Mass defect

\Delta m = \left[Zm_p + (A-Z)m_n\right] - M_{nucleus}

By Einstein’s mass–energy equivalence, this missing mass is exactly the energy that was released when the nucleons bound together, and so is also the energy you’d need to supply to pull the nucleus back apart, the binding energy:

Binding energy

\Delta E_b = \Delta m\, c^2

Worked example

Binding energy of the deuteron

A deuteron (²H nucleus) has mass 2.014102 u. A free proton has mass 1.007825 u and a free neutron has mass 1.008665 u. Find the binding energy, using 1 u = 931.5 MeV/c².

\Delta m = (1.007825+1.008665) - 2.014102 = \textcolor{#e08a1e}{0.002388\text{ u}}

\Delta E_b = 0.002388\times931.5 \approx \textcolor{#e08a1e}{2.224\text{ MeV}}

13.3 Binding Energy per Nucleon

Dividing binding energy by $A$ gives the binding energy per nucleon, a measure of how tightly, on average, each nucleon is held. Plotted against $A$ , this curve rises steeply for the lightest nuclei, peaks around $A\approx56$ (near iron), and then declines slowly for heavier nuclei.

This single curve explains why both fission and fusion release energy, even though they move in opposite directions:

13.3.1 Fission (splitting heavy nuclei)

A heavy nucleus near the low, right-hand end of the curve (such as uranium) splits into two medium-mass fragments sitting closer to the peak. Since the fragments are more tightly bound per nucleon than the original heavy nucleus was, the difference in binding energy is released, mostly as kinetic energy of the fragments.

13.3.2 Fusion (combining light nuclei)

Very light nuclei (such as hydrogen isotopes) sit on the steep, rising part of the curve. Fusing them into a slightly heavier nucleus moves the product further up the curve, toward the peak, so again binding energy per nucleon increases, and the difference is released as energy. This is why fusion, not fission, powers stars.

In both cases, the rule is the same: nuclei move toward the peak near $A\approx56$ , and energy is released whenever they do, whichever direction they approach it from.

13.4 Radioactive Decay Law

Radioactive decay of any one nucleus is random and unpredictable, but for a large collection of identical nuclei, the number decaying per unit time is proportional to the number of undecayed nuclei present, with a fixed per-nucleus decay probability per unit time, the decay constant $\lambda$ :

Decay rate

\dfrac{dN}{dt} = -\lambda N

Separating variables and integrating from $N_0$ at $t=0$ gives the exponential decay law:

Decay law

N = N_0e^{-\lambda t}

The half-life $T_{1/2}$ is the time for exactly half the sample to decay, $N/N_0=1/2$ at $t=T_{1/2}$ . Substituting and solving gives:

Half-life

T_{1/2} = \dfrac{\ln 2}{\lambda}

which lets the decay law be rewritten directly in terms of the half-life:

Decay law in terms of half-life

\dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^{t/T_{1/2}}

A related quantity is the mean life $\tau$ , the average lifetime of an individual nucleus before it decays:

Mean life

\tau = \dfrac{1}{\lambda} = \dfrac{T_{1/2}}{\ln 2} \approx 1.44\,T_{1/2}

13.4.1 Activity: how fast a sample is actually decaying

$dN/dt=-\lambda N$ is itself a rate, the number of decays happening per unit time, and that rate is given its own name, the activity $A$ of the sample:

Activity

\textcolor{#e08a1e}{A} = \left|\dfrac{dN}{dt}\right| = \lambda N

Since $N=N_0e^{-\lambda t}$ , activity decays by exactly the same exponential law as the nuclei count itself, $A=A_0e^{-\lambda t}$ , with the same half-life. The SI unit of activity is the becquerel ( $\text{Bq}$ ), one decay per second. An older unit, still common on sealed radioactive sources, is the curie ( $\text{Ci}$ ), originally defined as the activity of one gram of radium:

1\text{ Ci} = 3.7\times10^{10}\text{ Bq}

Activity, not the bare nuclei count $N$ , is what a Geiger counter actually measures and what radiation-safety limits are quoted in, since it is the decay rate, not the total amount of material, that determines how much radiation a sample emits per second.

Worked example

Activity of an iodine-131 sample

A sample contains $3.0\times10^{18}$ nuclei of iodine-131, half-life 8.0 days. Find its activity in both becquerel and curie.

\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{8.0\times86400} \approx 1.003\times10^{-6}\text{ s}^{-1}

A = \lambda N = (1.003\times10^{-6})(3.0\times10^{18}) \approx \textcolor{#e08a1e}{3.01\times10^{12}\text{ Bq}}

A = \dfrac{3.01\times10^{12}}{3.7\times10^{10}} \approx \textcolor{#e08a1e}{81.3\text{ Ci}}

The crucial point, easy to lose under exam pressure: $\lambda$ , $T_{1/2}$ , and $\tau$ are all fixed constants of the nuclide. None of them depend on how much sample you have, or how much time has already passed. A sample that has already gone through three half-lives decays exactly like a fresh sample, just starting from a smaller $N$ . See the interactive decay simulation to watch this play out.

Worked example

Fraction remaining after a given time

A radioactive sample has a half-life of 3 years. What fraction of the original sample remains after 12 years?

12 years is exactly 4 half-lives ( $12/3=4$ ):

\dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^4 = \textcolor{#e08a1e}{\dfrac{1}{16} = 6.25\%}

13.5 Types of Radioactive Decay

13.5.1 Alpha (α) decay

The nucleus emits an alpha particle, a helium nucleus ( $^4_2\text{He}$ , 2 protons + 2 neutrons bound together). Mass number falls by 4 and atomic number falls by 2:

^A_ZX \rightarrow\ ^{A-4}_{Z-2}Y + {}^4_2\text{He}

13.5.2 Beta (β) decay

A neutron inside the nucleus converts into a proton, emitting an electron (and an antineutrino): mass number stays the same, but atomic number rises by 1 (β⁻ decay, the common case). In the rarer β⁺ decay, a proton converts into a neutron, emitting a positron, and atomic number falls by 1 instead.

n \rightarrow p + e^- + \bar{\nu}

13.5.3 Gamma (γ) decay

No particles are ejected and neither $A$ nor $Z$ changes. A nucleus left in an excited state after an earlier alpha or beta decay simply drops to a lower energy state, emitting the excess energy as a high-energy photon, a gamma ray.

— From the NCERT exercises

Two of the chapter’s own exercise questions, with original worked solutions.

NCERT Exercise 13.1

Obtain the binding energy (in MeV) of a nitrogen nucleus (¹⁴₇N), given m(¹⁴₇N) = 14.00307 u. (mₕ = 1.007825 u, mₙ = 1.008665 u, 1 u = 931.5 MeV/c².)

Solution

Nitrogen has $Z=7$ protons and $A-Z=7$ neutrons. The mass defect, from §13.2:

\Delta m = 7(1.007825) + 7(1.008665) - 14.00307 = 14.11543 - 14.00307 = 0.11236\text{ u}

E_b = \Delta m \times 931.5 = 0.11236\times931.5 \approx \textcolor{#e08a1e}{104.7\text{ MeV}}

NCERT Exercise 13.9

Calculate the height of the potential barrier for a head-on collision of two deuterons. (Hint: the barrier height is the Coulomb repulsion between the two deuterons when they just touch, each taken as a hard sphere of radius 2.0 fm.)

Solution

Two deuterons, each of radius 2.0 fm, touch when their centres are a distance $d = 2\times2.0 = 4.0\text{ fm}$ apart. Each carries charge $+e$ (a deuteron is one proton plus one neutron), so by Coulomb’s law, exactly the kind of repulsion §13.1.1 says the nuclear force must overcome at short range:

U = \dfrac{ke^2}{d} = \dfrac{(9\times10^9)(1.6\times10^{-19})^2}{4.0\times10^{-15}} \approx 5.76\times10^{-14}\text{ J}

U = \dfrac{5.76\times10^{-14}}{1.6\times10^{-19}} \approx \textcolor{#e08a1e}{3.6\times10^5\text{ eV} = 360\text{ keV}}

This is why thermonuclear fusion needs temperatures of millions of degrees: ordinary thermal energies fall far short of the energy needed to push two deuterons close enough for the short-ranged nuclear force (§13.1.1) to take over from here.