Chapter 12 Nuclear Physics

Question 1

Einstein showed in his theory of relativity that matter can be considered a form of energy and hence, he proposed:

Answer 1

• Mass can be converted into energy
• Energy can be converted into mass

Question 2

• This is known as mass-energy equivalence, and can be summarised by the equation:

Answer 2

E = mc²

• Where:
  
  • E = energy (J)
  • m = mass (kg)
  • c = the speed of light (m s^-1)

Question 3

Some examples of mass-energy equivalence are:

Answer 3

• The fusion of hydrogen into helium in the centre of the sun
• The fission of uranium in nuclear power plants
• Nuclear weapons
• High-energy particle collisions in particle accelerators

Question 4

• Experiments into nuclear structure have found that the total mass of a nucleus is less than the sum of the masses of its constituent nucleons
• This difference in mass is known as the

Answer 4

mass defect

Question 5

• Mass defect is defined as:

Answer 5

The difference between an atom’s mass and the sum of the masses of its protons and neutrons

• The mass defect Δm of a nucleus can be calculated using:

Δm = Zm_p + (A – Z)m_n – m_total

• Where:
  
  • Z = proton number
  • A = nucleon number
  • m_p = mass of a proton (kg)
  • m_n = mass of a neutron (kg)
  • m_total = measured mass of the nucleus (kg)

Question 6

A system of separated nucleons has a greater mass than a system of bound nucleons

Answer 6

Question 7

Due to the equivalence of mass and energy, this decrease in mass implies that energy is

Answer 7

• released in the process
• Since nuclei are made up of neutrons and protons, there are forces of repulsion between the positive protons
  
  • Therefore, it takes energy, ie. the binding energy, to hold nucleons together as a nucleus

Question 8

• Binding energy is defined as:

Answer 8

The energy required to break a nucleus into its constituent protons and neutrons

• Energy and mass are proportional, so, the total energy of a nucleus is less than the sum of the energies of its constituent nucleons

Question 9

The formation of a nucleus from a system of isolated protons and neutrons is therefore an

Answer 9

• exothermic reaction - meaning that it releases energy
• This can be calculated using the equation:

E = Δmc²

Question 10

Binding Energy per Nucleon

Answer 10

• In order to compare nuclear stability, it is more useful to look at the binding energy per nucleon

Question 11

• The binding energy per nucleon is defined as:

Answer 11

The binding energy of a nucleus divided by the number of nucleons in the nucleus

Question 12

A higher binding energy per nucleon indicates a

Answer 12

• higher stability
  
  • In other words, it requires more energy to pull the nucleus apart
• Iron (A = 56) has the highest binding energy per nucleon, which makes it the most stable of all the elements

Question 13

Key Features of the Graph

• At low values of A

Answer 13

• Nuclei tend to have a lower binding energy per nucleon, hence, they are generally less stable
• This means the lightest elements have weaker electrostatic forces and are the most likely to undergo fusion

Question 14

Key Features of the Graph: Helium (⁴He), carbon (¹²C) and oxygen (¹⁶O) do not fit the trend

Answer 14

• Helium-4 is a particularly stable nucleus hence it has a high binding energy per nucleon
• Carbon-12 and oxygen-16 can be considered to be three and four helium nuclei, respectively, bound together

Question 15

Key Features of the Graph:

At high values of A:

Answer 15

• The general binding energy per nucleon is high and gradually decreases with A
• This means the heaviest elements are the most unstable and likely to undergo fission

Question 16

• Fusion is defined as:

Answer 16

The fusing together of two small nuclei to produce a larger nucleus

• Low mass nuclei (such as hydrogen and helium) can undergo fusion and release energy

Question 17

For two nuclei to fuse, both nuclei must have

Answer 17

high kinetic energy

• This is because the protons inside the nuclei are positively charged, which means that they repel one another

Question 18

It takes a great deal of energy to overcome the electrostatic force, so this is why it is can only be achieved in an

Answer 18

• extremely high-energy environment, such as star’s core
• When two protons fuse, the element deuterium is produced
• In the centre of stars, the deuterium combines with a tritium nucleus to form a helium nucleus, plus the release of energy, which provides fuel for the star to continue burning

Question 19

• Fission is defined as:

Answer 19

The splitting of a large atomic nucleus into smaller nuclei

• High mass nuclei (such as uranium) can undergo fission and release energy

Question 20

Fission must first be induced by

Answer 20

• firing neutrons at a nucleus
• When the nucleus is struck by a neutron, it splits into two, or more, daughter nuclei
• During fission, neutrons are ejected from the nucleus, which in turn, can collide with other nuclei which triggers a cascade effect
• This leads to a chain reaction which lasts until all of the material has undergone fission, or the reaction is halted by a moderator
• Nuclear fission is the process which produces energy in nuclear power stations, where it is well controlled

Question 21

When nuclear fission is not controlled, the chain reaction can cascade to produce the effects of a

Answer 21

nuclear bomb

Question 22

Significance of Binding Energy per Nucleon

• At low values of A:

Answer 22

• Attractive nuclear forces between nucleons dominate over repulsive electrostatic forces between protons
• In the right conditions, nuclei undergo fusion
• In fusion, the mass of the nucleus that is created is slightly less than the total mass of the original nuclei
  
  • The mass defect is equal to the binding energy that is released, since the nucleus that is formed is more stable

Question 23

Significance of Binding Energy per Nucleon

• At high values of A

Answer 23

• Repulsive electrostatic forces between forces begin to dominate, and these forces tend to break apart the nucleus rather than hold it together
• In the right conditions, nuclei undergo fission
• In fission, an unstable nucleus is converted into more stable nuclei with a smaller total mass
  
  • This difference in mass, the mass defect, is equal to the binding energy that is released

Question 24

Calculating Energy Released in Nuclear Reactions

Answer 24

The binding energy is equal to the amount of energy released in forming the nucleus

Question 25

The binding energy is equal to the amount of energy released in forming the nucleus, and can be calculated using:

Answer 25

**E = (Δm)c²** * Where: * E = Binding energy released (J) * Δm = **mass defect** (kg) * c = **speed of light** (m s^-1) * The daughter nuclei produced as a result of both fission and fusion have a higher binding energy per nucleon than the parent nuclei * Therefore, energy is released as a result of the mass difference between the parent nuclei and the daughter nuclei

Question 26

* Radioactive decay is defined as:

Answer 26

**The spontaneous disintegration of a nucleus to form a more stable nucleus, resulting in the emission of an alpha, beta or gamma particle**

Question 27

The random nature of radioactive decay can be demonstrated by observing the count rate of a Geiger-Muller (GM) tube

Answer 27

* When a GM tube is placed near a radioactive source, the counts are found to be irregular and cannot be predicted * Each count represents a decay of an unstable nucleus * These fluctuations in count rate on the GM tube **provide evidence for the randomness of radioactive decay**

Question 28

* A spontaneous process is defined as:

Answer 28

**A process which cannot be influenced by environmental factors**

Question 29

This means radioactive decay cannot be affected by environmental factors such as:

Answer 29

* Temperature * Pressure * Chemical conditions

Question 30

* A random process is defined as:

Answer 30

**A process in which the exact time of decay of a nucleus cannot be predicted** * Instead, the nucleus has a constant probability, ie. the same chance, of decaying in a given time * Therefore, with large numbers of nuclei, it is possible to statistically predict the behaviour of the entire group

Question 31

**average decay rate**

Answer 31

Since radioactive decay is spontaneous and random, it is useful to consider the average number of nuclei which are expected to decay per unit time

Question 32

* The decay constant λ is defined as:

Answer 32

**The probability that an individual nucleus will decay per unit of time**

Question 33

When a sample is highly radioactive, this means the number of decays per unit time is

Answer 33

* very high * This suggests it has a high level of **activity** * Activity, or the number of decays per unit time can be calculated using: ![]() * Where: * A = activity of the sample (Bq) * ΔN = number of decayed nuclei * Δt = time interval (s) * λ = decay constant (s^-1) * N = number of nuclei remaining in a sample

Question 34

The activity of a sample is measured in

Answer 34

**Becquerels** (Bq) * An activity of 1 Bq is equal to one decay per second, or 1 s^-1

Question 35

This equation shows:

Answer 35

* The greater the decay constant, the greater the activity of the sample * The activity depends on the number of undecayed nuclei remaining in the sample * The minus sign indicates that the number of nuclei remaining decreases with time - however, for calculations it can be omitted

Question 36

In radioactive decay, the number of nuclei falls very rapidly, without ever reaching zero * Such a model is known as

Answer 36

**exponential decay** * The graph of number of undecayed nuclei and time has a very distinctive shape

Question 37

* The number of undecayed nuclei N can be represented in exponential form by the equation:

Answer 37

**N = N₀e^–λt** * Where: * N₀ = the initial number of undecayed nuclei (when t = 0) * λ = decay constant (s^-1) * t = time interval (s)

Question 38

* The number of nuclei can be substituted for other quantities, for example, the activity A is directly proportional to N, so it can be represented in exponential form by the equation:

Answer 38

**A = A₀e^–λt**

Question 39

* The received count rate C is related to the activity of the sample, hence it can also be represented in exponential form by the equation:

Answer 39

**C = C₀e^–λt**

Question 40

**The exponential function** ***e***

Answer 40

* The symbol e represents the exponential constant * It is approximately equal to e = 2.718 * On a calculator it is shown by the button e^x * The inverse function of e^x is ln(y), known as the natural logarithmic function * This is because, if e^x = y, then x = ln(y)

Question 41

* Half life is defined as:

Answer 41

***The time taken for the initial number of nuclei to reduce by half*** * This means when a time equal to the half-life has passed, the activity of the sample will also half * This is because activity is proportional to the number of undecayed nuclei, A ∝ N

Question 42

***When a time equal to the half-life passes, the activity falls by half, when two half-lives pass, the activity falls by another half (which is a quarter of the initial value)***

Answer 42

Question 43

* The half life formula can then be derived as follows:

Answer 43

* Therefore, half-life t_½ can be calculated using the equation: ![]() * Where: * N = number of nuclei remaining in a sample * N₀ = the initial number of undecayed nuclei (when t = 0) * λ = decay constant (s^-1) * t = time interval (s) * * This equation shows that half-life t_½ and the radioactive decay rate constant λ are inversely proportional * Therefore, the shorter the half-life, the larger the decay constant and the **faster** the decay

Question 44

* The half life formula can then be derived as follows:

Answer 44

* Therefore, half-life t_½ can be calculated using the equation: * Where: * N = number of nuclei remaining in a sample * N₀ = the initial number of undecayed nuclei (when t = 0) * λ = decay constant (s^-1) * t = time interval (s) * This equation shows that half-life t_½ and the radioactive decay rate constant λ are inversely proportional * Therefore, the shorter the half-life, the larger the decay constant and the **faster** the decay