1. A nucleus is more than just mass
7.3 Nuclear Reactions
A nucleus is more than just mass

2. Objectives Define the unified mass unit;
State the meaning of the terms mass defect and binding energy and solve related problems;
Write nuclear reaction equations and balance the atomic and mass numbers;
Understand the meaning if the graph of binding energy per nucleon versus mass number;
State the meaning of and difference between fission and fusion;
Understand that nuclear fusion takes place in the core of stars;
Solve problems of fission and fusion reactions.

3. The unified mass unit Abbreviation: u or amu (atomic mass unit)
Definition: 1/12 the mass of an atom of carbon-12
A mole of 𝐶 is 12 g and the number of molecules is the Avogadro constant
An atom of carbon-12 has a mass of 6.02 x1023 x M = 12 g
M = 1.99 x kg
Hence M/12 = 1.66 x kg

4. Example
Find in units of u the masses of the proton, neutron, or electron.
Unified Mass Unit
x kg
1 u
electron
x kg
proton
x kg
neutron
x kg

5. The mass defect and binding energy
To find the mass of a nucleus: Mnucleus = Matom – Zme
Matom given in periodic table
For helium: Mnucleus = – 2 x = u
Lets check it by recalling that a helium nucleus is made up of 2 protons and 2 neutrons and adding their masses: 2 mp + 2 mn = u
But that’s larger than the mass of the nucleus by u! This leads to the concept mass defect.

6. The mass defect and binding energy
The mass of the protons plus neutrons is larger than the mass of the nucleus. This difference is defined as mass defect: δ = total mass of nucleons – mass of nucleus = Zmp + (A-Z)mn - Mnucleus

7. Example
Find the mass defect of the nucleus of gold, 𝐴𝑢 .

8. Einstein’s mass-energy formula
Where is the missing mass?
Einstein’s theory of special relativity states that mass and energy are equivalent and can be converted into each other: E = mc2
The mass defect of a nucleus has been converted into energy and is stored in the nucleus. This energy is called the binding energy of the nucleus and is denoted by Eb
Thus Eb = δc2
Binding energy = the work (energy) required to completely separate the nucleons of a nucleus
higher binding energy, more stability

9. Einstein’s mass-energy formula
How much energy corresponds to 1 u?
1 u x c2 = x x ( x 108)2 J
= x J
= MeV

10. Example
Find the energy equivalent to the mass of the proton, neutron, and electron.

11. Example
Find the binding energy of the nucleus of carbon-12.

12. The binding energy curve
We saw that the mass defect of helium is which corresponds to a binding energy of x MeV = MeV
The alpha particle has an unusually high stability which is why unstable particles decay into them
There are 4 nucleons in the helium nucleus so the binding energy per nucleon is MeV/4 = 7.1 MeV
Most nuclei have a binding energy per nucleon of approx. 8 MeV

13. The Binding Energy Curve

14. The binding energy curve
Maximum: 62 (Nickel)
If a nucleus heavier than this splits into two lighter ones, or if two lighter nuclei fuse together, then energy is released as a result.

15. Energy released in a decay
Consider the alpha decay of radium: 𝑅𝑎 → 𝑅𝑛 𝛼
For any decay, the total energy to the left of the arrow must equal the total decay on the right. Here, that energy corresponds to each mass according to Einstein’s formula + the kinetic energies of each mass
If the decaying radius is at rest, then the total energy available is Mc2 where M is the mass of the radium nucleus. To the right, we have the energies of the masses of the radon and the helium nuclei. Both have KE too.

16. Energy released in a decay
Thus, to be possible, the decay must be such that at the very minimum the energy corresponding to the radium mass is larger than the energies corresponding to the radon plus alpha particle masses.
Remember that your periodic table gives atomic, not nuclear, masses. You need to subtract the mass of the electrons…unless you’re only interested in mass differences (which we are), then atomic masses are ok.

17. Energy released in a decay
We see that the mass of radium exceeds that of radon + helium by u. Thus, there is kinetic energy released equal to x MeV = 4.84 MeV
Mass of radium = u Mass of radon = u + Mass of helium = u sum = u
If 50 g of radium were to decay in this way, the total energy released in this way would be N x 4.84 MeV where N is the total number of nuclei in the 50 g of radium. (1.3 x 1023).
The momenta of Rn and He are in opposite directions with equal magnitudes (momentum conservation). Vhelium ≈ 55 Vradon

18. In Summary
For the decay to take place the mass of the decaying nucleus has to be greater than the combined masses of the products (binding energy of the decaying nucleus must be less than the binding energies of the product nuclei.
This is why radioactive decay is possible for heavy elements lying to the right of nickel in the binding energy curve.

19. Assignment
Questions 1,2,3,5,6

20. Nuclear reactions
If a nucleus cannot decay by itself, it can still do so if energy is supplied to it by a fast moving particle that collides with it
Transmutation of nitrogen: an alpha particle colliding with nitrogen produces oxygen and hydrogen (a proton)
7 14 𝑁 𝛼 → 8 16 𝑂 𝑝
Atomic and mass numbers on both sides match
Left mass = u, right mass = u (larger) so the alpha must have enough kinetic energy to make up for the mass imbalance. KE>δ because products will also have KE

21. Nuclear reactions
Generally, in a reaction 𝐴+𝐵→𝐶+𝐷, energy will be released if ∆𝑚 → 𝑚 𝐴 + 𝑚 𝐵 − 𝑚 𝐶 + 𝑚 𝐷 is positive.
The amount of energy released is equal to 𝐸=∆𝑚 𝑐 2
2 kinds of energy producing nuclear reactions: Fission & Fusion

22. Nuclear Fission A heavy nucleus splits up into lighter nuclei
Uranium-235 absorbs a neutron, momentarily becoming uranium-236, which then splits into lighter nuclei.
Many possibilities for daughter nuclei
𝐵𝑎 𝐾𝑟 𝑛
Production of neutrons is characteristic of fission; they can cause other fission reactions – chain reaction
Minimum mass of uranium- 235 is needed, otherwise, neutrons escape without causing further reactions – critical mass

23. Nuclear fission The energy released can be calculated as follows:
This energy appears as the kinetic energy of the products
This is LOTS of energy (1 kg = 7 x 1013 J)
Mass of uranium plus neutron
= u
Mass of products
= u u + 3 x u
= u
Mass difference
= u – u = u
Energy released
= x MeV = MeV

24. Nuclear fission Nuclear reactors – controlled rate of reaction
Nuclear bomb – too much energy produced at 1 time – uncontrolled rate of reaction
Alpha particles cannot be used to start fission because protons would be repelled by the uranium nucleus
An electron would not sufficiently perturb the heavy nucleus.
MUST be a neutron

25. Nuclear Fusion
The joining of 2 light nuclei into a heavier one + energy
1 2 𝐻 𝐻 → 2 3 𝐻𝑒 𝑛
1 kg deuterium ≈ 1013 J
2 x mass of deuterium
= u
Mass of helium + neutron
= u
Mass difference
= u – u = u
Energy released
= x MeV = 3.26 MeV

26. Mass of helium + positrons + neutrinos
Example
Another fusion reaction is 𝐻 → 2 4 𝐻𝑒 𝑒 +2 ν 𝑒 𝛾 , where 4 h nuclei fuse into a He plus 2 positrons (electron antiparticles – same mass, opposite charge), 2 electron neutrinos and a photon. Calculate the energy released in this reaction.
4 x mass of hydrogen
= u
Mass of helium + positrons + neutrinos
= ( – 2 x ) + (2 x )
= u
Mass difference
= u
Energy released
= x MeV
= 24.7 MeV

27. Nuclear Fusion
For light nuclei to fuse, very high temperatures are required to overcome the electrostatic repulsion between nuclei.
The enormous temperature causes the nuclei to move fast enough to approach each other sufficiently for fusion to occur.
Temperatures > 107 K = plasma state (ionized atoms)
Plasma cannot contact other things – must be contained by magnetic fields (tokamaks)

28. Tokamaks
Serious unsolved problems with prolonged confinement of plasmas – not a viable source of commercial energy any time soon

29. Fusion in Stars
The high temperatures and pressures in the interior of stars make stars ideal places for fusion.
High temperature needed to overcome electric repulsion
High pressure ensures that sufficient numbers of nuclei are found close to each other, increasing probability of fusion
4 1 1 𝐻 → 2 4 𝐻𝑒 𝑒 +2 ν 𝑒 𝛾 - common reaction in stellar cores
Fusion = source of star’s energy, prevents it from collapsing under its own weight and provides energy that star sends out as light and heat.
Stars = element factories

30. Summary

31. Assignment
Questions: 7,8,9,10,12