Atoms and Nuclei • Topic 2 of 3

The Nucleus: Composition, Size & Binding Energy

The nucleus is the tiny, dense core of the atom. It is made of nucleonsprotons (charge $+e$, mass $\approx 1.0073\ \text{u}$) and neutrons (no charge, mass $\approx 1.0087\ \text{u}$). The number of protons is the atomic number $Z$, the total number of nucleons is the mass number $A$, and the number of neutrons is $N = A - Z$. A nucleus is written ${}^{A}_{Z}X$.

  • Isotopes: same $Z$, different $A$ (e.g. ${}^{1}_{1}\text{H}$, ${}^{2}_{1}\text{H}$, ${}^{3}_{1}\text{H}$). Same chemistry, different mass.
  • Isobars: same $A$, different $Z$ (e.g. ${}^{3}_{1}\text{H}$ and ${}^{3}_{2}\text{He}$).
  • Isotones: same number of neutrons $N$ (e.g. ${}^{3}_{1}\text{H}$ and ${}^{4}_{2}\text{He}$, both with $N = 2$).

Experiments show the nuclear radius obeys $R = R_0 A^{1/3}$, where $R_0 = 1.2\ \text{fm} = 1.2 \times 10^{-15}\ \text{m}$. Because volume $\propto R^3 \propto A$, the nuclear volume is proportional to the number of nucleons. This means the nuclear density is essentially the same for all nuclei, an enormous $\rho \approx 2.3 \times 10^{17}\ \text{kg m}^{-3}$ — about $10^{14}$ times the density of water.

A startling fact emerges when nucleons bind: the mass of a nucleus is always less than the sum of the masses of its free nucleons. This shortfall is the mass defect $\Delta m = [Z m_p + (A - Z) m_n] - M_{\text{nucleus}}$. By Einstein's mass-energy relation $E = \Delta m\, c^2$, the missing mass appears as energy released when the nucleus forms. A convenient conversion is $1\ \text{u} = 931.5\ \text{MeV}/c^2$, so $1\ \text{u}$ of mass defect corresponds to $931.5\ \text{MeV}$.

The energy equivalent of the mass defect is the binding energy $E_b = \Delta m\, c^2$ — the energy needed to split the nucleus into its separate nucleons, or equivalently the energy released when it is assembled. More useful for comparing nuclei is the binding energy per nucleon, $\frac{E_b}{A}$, which measures how tightly each nucleon is held.

Plotting $\frac{E_b}{A}$ against $A$ gives the famous binding-energy curve. It rises steeply for light nuclei, reaches a broad peak of about $8.8\ \text{MeV}$ near iron-56, then falls slowly for heavy nuclei to about $7.6\ \text{MeV}$ for uranium. Two consequences follow: very heavy nuclei can release energy by splitting (fission) toward the peak, while very light nuclei can release energy by combining (fusion) toward the peak. Nuclei near iron are the most stable in the universe.

Binding energy per nucleon vs mass number A (peak near Fe-56)Mass number AE_b / A (MeV)HHeFe-56 (~8.8)U-238fusion →← fission
1
Worked Example
The radius of the ${}^{27}_{13}\text{Al}$ nucleus is measured. Find it, taking $R_0 = 1.2\ \text{fm}$.
Solution
  1. $R = R_0 A^{1/3}$ with $A = 27$.
  2. $A^{1/3} = 27^{1/3} = 3$.
  3. $R = 1.2 \times 3 = 3.6\ \text{fm} = 3.6 \times 10^{-15}\ \text{m}$.

Answer: The aluminium nucleus has radius $3.6\ \text{fm}$.

2
Worked Example
Find the ratio of the radii of the nuclei ${}^{27}_{13}\text{Al}$ and ${}^{125}_{52}\text{Te}$.
Solution
  1. $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{27}{125}\right)^{1/3}$.
  2. $= \frac{27^{1/3}}{125^{1/3}} = \frac{3}{5}$.

Answer: $R_1 : R_2 = 3 : 5$.

3
Worked Example
Calculate the mass defect and binding energy of the helium nucleus ${}^{4}_{2}\text{He}$. Take $m_p = 1.00728\ \text{u}$, $m_n = 1.00866\ \text{u}$, $M(\text{He}) = 4.00260\ \text{u}$.
Solution
  1. Sum of nucleon masses $= 2(1.00728) + 2(1.00866) = 2.01456 + 2.01732 = 4.03188\ \text{u}$.
  2. Mass defect $\Delta m = 4.03188 - 4.00260 = 0.02928\ \text{u}$.
  3. Binding energy $E_b = 0.02928 \times 931.5 = 27.3\ \text{MeV}$.

Answer: $\Delta m = 0.02928\ \text{u}$ and $E_b \approx 27.3\ \text{MeV}$.

4
Worked Example
Find the binding energy per nucleon for the helium nucleus of Example 3.
Solution
  1. Binding energy $E_b = 27.3\ \text{MeV}$, nucleon number $A = 4$.
  2. $\frac{E_b}{A} = \frac{27.3}{4}$.
  3. $= 6.8\ \text{MeV per nucleon}$.

Answer: The binding energy per nucleon of ${}^{4}_{2}\text{He}$ is about $6.8\ \text{MeV}$.

5
Worked Example
How much energy is equivalent to a mass of $1\ \text{u}$? Verify the standard value using $E = mc^2$.
Solution
  1. $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$.
  2. $E = mc^2 = 1.66 \times 10^{-27} \times (3 \times 10^8)^2 = 1.494 \times 10^{-10}\ \text{J}$.
  3. Convert: $E = \frac{1.494 \times 10^{-10}}{1.6 \times 10^{-13}} \approx 934\ \text{MeV} \approx 931.5\ \text{MeV}$.

Answer: $1\ \text{u} \equiv 931.5\ \text{MeV}$, confirming the standard conversion.

6
Worked Example
Why is the density of a nucleus independent of its mass number $A$?
Solution
  1. Nuclear mass $\approx A m_n$ (proportional to $A$).
  2. Nuclear volume $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A$ (also proportional to $A$).
  3. Density $\rho = \frac{\text{mass}}{\text{volume}} = \frac{A m_n}{\frac{4}{3}\pi R_0^3 A} = \frac{m_n}{\frac{4}{3}\pi R_0^3}$ — the $A$ cancels.

Answer: Since both mass and volume grow in direct proportion to $A$, their ratio is constant ($\rho \approx 2.3 \times 10^{17}\ \text{kg m}^{-3}$) for all nuclei.

Key Points

  • A nucleus ${}^{A}_{Z}X$ has $Z$ protons and $N = A - Z$ neutrons; isotopes share $Z$, isobars share $A$, isotones share $N$.
  • Nuclear radius $R = R_0 A^{1/3}$ with $R_0 = 1.2\ \text{fm}$; volume $\propto A$.
  • Nuclear density is the same for all nuclei, about $2.3 \times 10^{17}\ \text{kg m}^{-3}$.
  • Mass defect $\Delta m = [Z m_p + (A - Z) m_n] - M_{\text{nucleus}}$ gives binding energy $E_b = \Delta m\, c^2$, with $1\ \text{u} = 931.5\ \text{MeV}$.
  • Binding energy per nucleon $\frac{E_b}{A}$ peaks near $8.8\ \text{MeV}$ at iron-56, allowing energy release by fusion (light nuclei) and fission (heavy nuclei).
Tap an option to check your answer0 / 4
Q1.Nuclei having the same mass number $A$ but different atomic number $Z$ are called:
Explanation: Same $A$, different $Z$ defines isobars (e.g. ${}^{3}_{1}\text{H}$ and ${}^{3}_{2}\text{He}$).
Q2.The nuclear radius is related to the mass number by:
Explanation: Experimentally $R = R_0 A^{1/3}$ with $R_0 = 1.2\ \text{fm}$.
Q3.The mass of a nucleus compared with the sum of the masses of its nucleons is:
Explanation: The nucleus is lighter by the mass defect $\Delta m$, whose energy equivalent is the binding energy.
Q4.On the binding-energy-per-nucleon curve, the maximum stability occurs near:
Explanation: The curve peaks at about $8.8\ \text{MeV per nucleon}$ near iron-56.