Grade 12

Advanced

8 min

Lesson 8 of 12

Nuclear Structure and Binding Energy

Not Started

Analyzing the forces that hold the nucleus together and the 'missing mass' that converts into energy.

If you weighed the individual bricks of a Lego castle and then weighed the completed castle, you'd expect them to be the same—but in the atomic world, the 'castle' is actually lighter than its parts. Where does that missing mass go, and how does it keep the universe from flying apart?

Learning Objectives

1.Define mass defect and calculate it using the atomic mass unit ( $u$ )
2.Calculate the binding energy per nucleon to determine the relative stability of isotopes
3.Explain how the strong nuclear force overcomes electrostatic repulsion within the nucleus

The Battle Inside the Nucleus

Inside every atom, a violent struggle occurs. Protons are positively charged and packed into a space of roughly $10^{-15}$ meters. According to Coulomb's Law, these protons should repel each other with immense force. What prevents the nucleus from exploding? The Strong Nuclear Force. This force is approximately 100 times stronger than electromagnetism but has an incredibly short range. It acts as the 'nuclear glue' that binds protons and neutrons (nucleons) together. Interestingly, this force is charge-independent, meaning it attracts $p-p$ , $n-n$ , and $p-n$ with equal intensity, provided they are close enough to touch.

Before calculating mass defect, you must identify the number of nucleons. For an isotope of Carbon-14 ( $^{14}_{6}C$ ): 1. Identify the atomic number ( $Z$ ): $Z = 6$ (protons). 2. Identify the mass number ( $A$ ): $A = 14$ (total nucleons). 3. Calculate neutrons ( $N$ ): $N = A - Z = 14 - 6 = 8$ neutrons.

Quick Check

If the strong nuclear force is so much stronger than the electrostatic force, why don't all the atoms in the universe clump together into one giant nucleus?

Think about the distance over which these forces operate.

Answer

The strong nuclear force has an extremely short range (about $10^{-15}$ m); it cannot pull on nucleons that are not immediate neighbors.

The Mystery of Mass Defect

When nucleons assemble into a nucleus, the total mass of the final nucleus is always less than the sum of the individual masses of its protons and neutrons. This difference is the mass defect (

\Delta m

). Where does the mass go? Einstein’s mass-energy equivalence,

E = mc^2

, provides the answer: the 'missing' mass is converted into energy and released. This is the Binding Energy (

BE

). The more mass that is 'lost,' the more energy is released, and the more tightly the nucleus is held together. To calculate it, we use:

\Delta m = [Z(m_p) + N(m_n)] - m_{nucleus}

Calculate the mass defect for Helium-4 ( $^{4}_{2}He$ ). Given: $m_p = 1.007276 \text{ u}$ , $m_n = 1.008665 \text{ u}$ , and the measured mass of Helium-4 is $4.001506 \text{ u}$ . 1. Sum of parts: $2(1.007276) + 2(1.008665) = 4.031882 \text{ u}$ . 2. Subtract measured mass: $4.031882 - 4.001506 = 0.030376 \text{ u}$ . 3. The mass defect $\Delta m = 0.030376 \text{ u}$ .

Quick Check

True or False: A nucleus always weighs less than the sum of the individual protons and neutrons that compose it.

Answer

True. This 'missing mass' is the mass defect, which was released as binding energy during the formation of the nucleus.

Binding Energy per Nucleon

Total binding energy doesn't tell the whole story of stability. A large nucleus like Uranium has a huge total $BE$ , but it is actually quite unstable. To measure stability, we look at Binding Energy per Nucleon ( $BE/A$ ). This is the average energy required to remove a single nucleon from the nucleus. The higher the $BE/A$ , the more stable the atom. The 'Goldilocks' of stability is Iron-56 ( $^{56}Fe$ ), which has the highest $BE/A$ . Elements lighter than Iron tend to undergo fusion to reach that peak, while heavier elements undergo fission to move toward it.

Calculate the $BE/A$ for Iron-56 ( $^{56}_{26}Fe$ ) given $\Delta m = 0.528489 \text{ u}$ . 1. Convert mass defect to energy using $1 \text{ u} = 931.5 \text{ MeV}$ : $BE = 0.528489 \times 931.5 = 492.28 \text{ MeV}$ . 2. Divide by the number of nucleons ( $A = 56$ ): $BE/A = 492.28 / 56 = 8.79 \text{ MeV/nucleon}$ . 3. Compare: Since Uranium-235 has a $BE/A$ of only $\approx 7.6 \text{ MeV}$ , Iron-56 is significantly more stable.

Key Takeaways

1The Strong Nuclear Force binds nucleons together, overcoming the electrostatic repulsion of protons over very short distances.
2Mass defect ( $\Delta m$ ) is the difference between the mass of the individual nucleons and the actual mass of the nucleus.
3Binding energy ( $E = \Delta m c^2$ ) is the energy released when a nucleus forms; it is also the energy required to break it apart.
4Binding energy per nucleon ( $BE/A$ ) is the true indicator of nuclear stability, peaking at Iron-56.

Quick Check

Multiple Choice

What happens to the 'missing mass' in the mass defect?

Multiple Choice

Which of the following isotopes is the most stable?

True/False

The strong nuclear force is effective over long distances, similar to gravity.

What's Next?

Review Tomorrow

In 24 hours, try to explain to someone why a nucleus is lighter than its parts and how that relates to Einstein's most famous equation.

Practice Activity

Find the mass of a Carbon-12 nucleus and the masses of 6 protons and 6 neutrons. Calculate the mass defect in atomic mass units ( $u$ ).

Browse

Browse

Nuclear Structure and Binding Energy

Learning Objectives

The Battle Inside the Nucleus

The Mystery of Mass Defect

Binding Energy per Nucleon

Key Takeaways

Quick Check

What's Next?

Nuclear Structure and Binding Energy

Learning Objectives

The Battle Inside the Nucleus

The Mystery of Mass Defect

Binding Energy per Nucleon

Key Takeaways

Quick Check

What's Next?