### Collection Nuclear stability

## Nuclear stability

Lesson for 16-19

Why are some nuclides stable while others are unstable (radioactive)? Starting from the pattern of stability, this section looks at forces in the nucleus and the idea of binding energy.

## Episode 523: Preparation for nuclear stability topic

Teaching Guidance for 16-19

Starting from the pattern of stability, this section looks at forces in the nucleus and the idea of binding energy.

#### Main aims of this topic

Students will:

- Sketch the
*N*-*Z*graph for stable nuclei - Describe the balance of forces that results in a stable nucleus
- Calculate mass defect and binding energy
- Relate nuclear fission and fusion to the graph of binding energy per nucleon

#### Prior knowledge

Students should know about the composition of nuclei in terms of protons and neutrons. They should be familiar with nuclear notation e.g. 20983Bi

#### Where this leads

An understanding of nuclear stability leads on to a study of nuclear fusion and fission.

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### Stable nuclides

## Episode 524: Stable nuclides

Lesson for 16-19

- Activity time 65 minutes
- Level Advanced

Hundreds of nuclides are known; this episode looks at the pattern of stability using an N-Z plot, and considers the need for an additional, attractive force to hold the nucleus together.

Lesson Summary

- Discussion: Which stable nuclides exist? (10 minutes)
- Student Activity: Using Excel to generate an N-Z plot (20 minutes)
- Discussion: Interpreting the graph (5 minutes)
- Worked example: Calculating the force between protons (10 minutes)
- Discussion: The nature of the strong nuclear force (5 minutes)
- Worked example and student example: Calculating nuclear densities (15 minutes)

#### Discussion: Which stable nuclides exist

Rehearse and extend your students’ existing knowledge. Nuclei are composed of N neutrons and *Z* protons – collectively *A* = *N* + *Z* nucleons. Explain (or revise) nuclide notation; for example:

The heaviest stable element is bismuth-83, 20983Bi

How many protons and neutrons in this nucleus? (83 and 126)

There are 90 naturally-occurring elements between hydrogen ( *Z* is 1) to uranium ( *Z* is 92). Is there anything odd about this? (Two are missing

; these are purely artificial and man made, 9943Tc (technitium) and 6361Pm (promethium). Both are radioactive with relatively short half lives on a geological time scale, and thus would have decayed long ago.)

All elements beyond *Z* of 92 are man-made. So far (as of February 2004) the record is *Z* of 116.

#### Student activity: Using Excel to generate an *N*-*Z* plot

Use a spreadsheet to generate a plot of *N* versus *Z* for stable nuclides. To save time (and transcription errors!) you could prepare an Excel spreadsheet with the data below already entered (columns for the element name, *A*, *Z* and, anticipating work to follow, the nucleus mass in atomic mass units *u* – see below).

Representative stable nuclides:

Episode 524-1: Stability: Balanced numbers of neutrons and protons (Word, 91 KB)

#### Discussion: Interpreting the graph

Describe the graph. (It is linear (*N* ∝ *Z*) up to *Z* ~ 20, then increasingly *N* > *Z*: there is a neutron excess

. The neutron excess is crucial in explaining nuclear stability, and for setting up a chain reaction in the exploitation of nuclear reactions.

Think about the Coulomb repulsion between protons. What are the neutrons doing in there? (Neutrons must help overcome the strong repulsion between the protons, partly by diluting them

, but also providing an attractive force to balance the electric repulsion. Hence the name strong nuclear force.

#### Worked examples: Calculating the force between protons

If Coulomb’s law has been covered, calculate the repulsive force between two protons that just touch

, so the separation of their centres r is the diameter of a proton 1.4 × 10^{-15} m).

*F*_{e} = 14 π *ε*_{0} *q*_{1} *q*_{2}*r*^{ 2}

where *q*_{1} and *q*_{2} both have the same value, +*e* = 1.6 × 10^{-19} C. So:

*F*_{e} ~ 100 N

#### Discussion: The nature of the strong nuclear force

Explain the need for the strong nuclear force to balance the Coulomb repulsion. This force must be attractive – it overcomes the coulomb repulsion; independent of electric charge – it acts between nn, pp and np; and very short range ~ 1 fm = 1 × 10^{-15} m.

#### Worked example and student example: Calculating nuclear densities

Calculate the density of a nucleon:

nucleon mass ~1.7 × 10^{-27} kg; radius ~1.4 × 10^{-15} m; density r ~1.4 × 10^{17} kg m^{-3}

This is enormous compared to the density of everyday matter. Ask your students to repeat the calculation for He-4 (or give data for other nuclides):

nuclide mass ~6.6 × 10^{-27} kg; radius ~2 × 10^{-15} m; density ~1.6 × 10^{17} kg^{-3}

So the density is roughly the same for all nuclei. This is summed up in the relationship:

*r* = *r*_{0} × *A*^{ ⅓ }

with *r*_{0} = 1.4 fm

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### Binding energy

## Episode 525: Binding energy

Lesson for 16-19

- Activity time 100 minutes
- Level Advanced

This episode discusses: mass defect and atomic mass units; and, fission and fusion linked to binding energy. And is accompanied by worked examples, student question sets, and student activities.

Lesson Summary

- Discussion: Introducing mass defect and atomic mass units (10 minutes)
- Discussion: Mass defect and binding energy (10 minutes)
- Worked example: Calculating binding energy (10 minutes)
- Student questions: Calculations (20 minutes)
- Student activity: Spreadsheet calculations (20 minutes)
- Student activity: Spreadsheet calculations of binding energy per nucleon (20 minutes)
- Discussion: Fission and fusion linked to binding energy graph (10 minutes)

#### Discussion: Introducing mass defect and atomic mass units

Ask your students to consider whether the following data is self-consistent:

proton mass, m_{p} = 1.673 × 10^{-27} kg

neutron mass, m_{n} = 1.675 × 10^{-27} kg

mass of a 42He nucleus = 6.643 × 10^{-27} kg

The mass of a 42He nucleus is *less* than the sum of the masses of its parts; this is true for *all* nuclides. So much for conservation of mass.

Introduce the atomic mass unit (amu, or *u*) as a convenient unit of nuclear mass. 1 amu or 1 *u*, which is 112 the mass of a neutral 126C atom (i.e. including its six electrons):

*u* = 1.66056 × 10^{-27} kg.

Thus:

m_{p} = 1.0073 *u*

m_{n} = 1.0087 *u*

m_{e} = 0.00055 *u*

mass of a neutral

42He

atom = 4.0026 u

#### Discussion: Mass defect and binding energy

What has happened to the missing mass – or mass defect – between the whole and the sum of the parts? To separate the nucleons of a nucleus, work has to be done against the attractive strong nuclear force. The work done is known as the binding energy. If the nucleons are not bound, then it takes no energy to separate them; the binding energy when they are separated is zero. If protons and neutrons (nucleons) are bound together in a nucleus, the bound nucleus must have *less* rest energy than the rest energy of the nucleons of which it is made.

In turn, this means that the mass of the nucleus must be less than the sum of the masses of its nucleons.

Binding energy can be a rather confusing term because students often think that this means that energy is required to bind nucleons together. As with chemical bonds, this is the opposite of the truth. Energy is needed to break bonds, or separate nucleons.

Einstein’s Special Theory of Relativity (1905) relates mass and energy via the equation *E* = *m* × *c*^{ 2} (where *c* is the speed of light in a vacuum). In this case, we have:

binding energy = mass defect × *c*^{ 2}

or

Δ *E* = Δ *m* × *c*^{ 2}

(It is not advisable to talk about mass being converted to energy

or similar expressions. It is better to say that, in measuring an object’s mass, we are determining its energy. A helium nucleus has less mass than its constituent nucleons; in pulling them apart, we do work and so give them energy; hence their mass is greater.)

#### Worked examples: Calculating binding energy

Calculate the mass defect and binding energy for 42He

(Mass defect = 0.053 × 10^{-27} kg;

binding energy = 1.59 × 10^{-12} J

binding energy = 9.94 MeV)

#### Student questions

Episode 525-1: Change in energy: Change in mass (Word, 189 KB)

Episode 525-2: Finding binding energy (Word, 52 KB)

Episode 525-3: Fusion in a kettle? (Word, 35 KB)

#### Student activity: A data analysis exercise using Excel

This uses a spreadsheet to calculate binding energy for a number of nuclides.

Episode 525-4: A binding energy calculator (Word, 130 KB)

#### Student activity

Another spreadsheet activity, this time looking at the binding energy per nucleon. Note that it is desirable to plot this graph with a negative energy axis; this means that the lowest values are for the most stable nuclides.

Episode 525-5: Binding energy of nuclei (Word, 79 KB)

#### Discussion

Briefly discuss fission and fusion in terms of the graph. Although the fission jump

looks quite small compared to a typical fusion jump, the graph is plotting BE *per nucleon*. Many more nucleons are involved in the fission of heavy atoms than in the fusion of lighter ones. (This topic can be developed further when discussing nuclear power.)

Episode 525-6: Binding energy per nucleon (Word, 45 KB)