
We infer the presence of a strong nuclear force because positively charged protons in the nucleus would otherwise repel each other.
The strong nuclear force is much stronger than the Coulomb force, but its range does not extend beyond the approximate size of a nucleus.
The energy captured in the nucleus is called binding energy and is calculated by Einstein’s equation of mass–energy equivalence.
The mass of a nucleus of an element is less than the mass of its constituent nucleons. The difference is called mass deficiency.
The mass deficiency and binding energy of a nucleus are related through Einstein’s equation.
For more information to research the historical development of the concepts of the strong (and weak) forces, see
 Coming of Age in the Milky Way by Timothy Ferris,
 Strange Beauty by George Johnson, and
 Story of the W and Z by Peter Watkins.

strong nuclear force, atomic mass unit (amu), mass–energy equivalence, rest energy, binding energy, mass deficiency



Review problems and questions 

 Research and describe the historical development of the concept of the strong nuclear force.

Answers will vary, but will likely include some of the following points: Rutherford’s scattering experiment showed that the nucleus was dense but also composed of many positively charged protons, which should repel each other,
 The neutron was discovered in 1932 by James Chadwick, indicating that the force holding the nucleus together must work for both positive particles (protons) and neutral particles (neutrons),
 Hideki Yukawa proposed a model for the nuclear force in 1934,
 The existence of quarks was proposed in 1964 by Murray GellMann and George Zweig as the particles whose interactions are at the heart of the strong nuclear force,
 Quarks were experimentally confirmed in 1968 at the Stanford Linear Accelerator Center.


What is the rest energy of an electron (whose mass = 9.11 × 10^{−31} kg)?
 1.0 × 10^{−47} J
 3.0 × 10^{−39} J
 2.7 × 10^{−22} J
 8.2 × 10^{−14} J

d is correct. Use the massenergy equivalence formula, E = mc^{2}:E = mc^{2} = (9.11 × 10^{−31} kg) × (3 × 10^{8} m/s)^{2} = 8.2 × 10^{−14} J


In your body, are there more protons than neutrons? More protons than electrons?

From the chart of nuclides we know that all isotopes have in general more neutrons than protons. Therefore in our body we have more neutrons than protons. Most of the atoms that make up our body are neutral, which means that the number of protons in our body is equal to the number of electrons.


The isotope ${}_{8}{}^{16}\text{O}$
has a mass of 15.99491 amu. What is its binding energy?

Answer: Binding energy per nucleon for oxygen16 is 7.99 MeV
Solution: Let’s start by collecting all the information that we know. The atomic mass of oxygen16 is m_{O} = 15.99491 amu.
 The atomic number of oxygen is Z = 8.
 Therefore, the neutron number is N = 8.
 The mass of a neutron is m_{n} = 1.00869 amu.
 The atomic mass of hydrogen is m_{H} = 1.007825 amu.
Now we calculate the mass deficiency (Δm) for the oxygen16 nucleus:$$\begin{array}{lll}\text{\Delta}m\hfill & =\hfill & 8{m}_{\text{H}}+8{m}_{\text{n}}{m}_{\text{O}}\hfill \\ \hfill & =\hfill & 8(1.007825\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{amu})+8(1.00869\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{amu})15.994915\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{amu}\hfill \\ \hfill & =\hfill & 0.137205\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{amu}\hfill \end{array}$$Since there are 931.5 MeV per amu, the total amount of binding energy associated with a nucleus of oxygen16 is (0.137205 amu) × (931.5 MeV/amu) = 127.8065 MeVThe binding energy of the oxygen16 nucleus per nucleon is therefore(127.8065 MeV)/(16 nucleons) = 7.99 MeV per nucleon

 Calculate the mass deficiency for the nucleus of tritium ${}_{1}{}^{3}\text{H}$, which is an isotope of hydrogen with a mass of 3.0160 amu. (The mass of a proton is 1.0073 amu, the mass of a neutron is 1.0087 amu, and the mass of an electron is 0.0005 amu.)

Answer: 0.0092 amu
The tritium atom is an isotope of hydrogen; therefore, its nucleus has one proton. Tritium has three nucleons, so the other two must be neutrons. The sum of the masses of the nucleons that make up the tritium nucleus is therefore$${m}_{\text{p}}+2{m}_{\text{n}}=1.0073\text{amu}+2\times 1.0087\text{amu}=3.0247\text{amu}$$The mass defect is the difference between the sum of the masses of the nucleons and the mass of the tritium nucleus (which is 3.0160 amu − 0.0005 amu = 3.0155 amu):$$\text{\Delta}m=3.0247\text{amu}3.0155\text{amu}=0.0092\text{amu}$$

 What is the binding energy of the nucleus of a tritium atom?

Answer: 1.36 × 10^{−12} J
The binding energy is the energy equivalent of the mass defect, which was calculated above to be 0.0092 amu. Use Einstein’s mass–energy equation and the conversion from atomic mass units to kilograms:$$E=m{c}^{2}=(0.0092\text{amu})\left(\frac{1.661\times {10}^{27}\text{kg}}{1\text{amu}}\right){\left(3\times {10}^{8}\text{m/s}\right)}^{2}=1.36\times {10}^{12}\text{J}$$

Take a Quiz 