Pearson Edexcel Physics

Unit 5: Thermodynamics, Radiation, Oscillations and Cosmology

5.4 Nuclear Decay

Pearson Edexcel Physics Unit 5: Thermodynamics, Radiation, Oscillations and Cosmology 5.4 Nuclear Decay Candidates will be assessed on their ability to:
133.	Understand the concept of nuclear binding energy and be able to use the equation [math]ΔE = c^2 Δm[/math] in calculations of nuclear mass (including mass deficit) and energy
134.	Use the atomic mass unit (u) to express small masses and convert between this and SI units
135.	Understand the processes of nuclear fusion and fission with reference to the binding energy per nucleon curve
136.	Understand the mechanism of nuclear fusion and the need for very high densities of matter and very high temperatures to bring about and maintain nuclear fusion
137.	Understand that there is background radiation and how to take appropriate account of it in calculations
138.	Understand the relationships between the nature, penetration, ionizing ability and range in different materials of nuclear radiations (alpha, beta and gamma)
139.	Be able to write and interpret nuclear equations given the relevant particle symbols
140.	CORE PRACTICAL 15: Investigate the absorption of gamma radiation by lead
141.	Understand the spontaneous and random nature of nuclear decay
142.	Be able to determine the half-lives of radioactive isotopes graphically and be able to use the equations for radioactive decay [math]\text{activity } A=\lambda N, \, \frac{dN}{dt}=-\lambda N, \, \lambda=\frac{\ln 2}{t_{1/2}}, \, N=N_0 e^{-\lambda t} \text{ and } A=A_0 e^{-\lambda t}[/math] Derive and use the corresponding log equations.

133) Understand the concept of nuclear binding energy and be able to use the equation [math]ΔE = c^2 Δm[/math] in calculations of nuclear mass (including mass deficit) and energy
134) Use the atomic mass unit (u) to express small masses and convert between this and SI units
⇒ Nuclear Binding Energy:
The mass deficit comes about because a small amount of the mass of the nucleons is converted into the energy needed to hold the nucleus together. This is called binding energy, [math][/math]. It is calculated using Einstein’s mass-energy relationship:
[math]ΔE = c^2 Δm[/math]
Where c is the speed of light.
There are two common systems of units for calculating binding energy. If you have calculated the mass deficit in kilograms.
(SI units) then using [math]c = 3.00 × 10^8 m.s^{-1}[/math] will give the binding energy in joules. Alternatively, if you have calculated the mass deficit in atomic mass units, then you convert this into binding energy in mega-electron volts (MeV) using:
[math]1u = 931.5 MeV[/math]
Example:
Calculate the binding energy for a carbon-12 nucleus in both joules and electron volts.
Working in kilograms
[math]\begin{gather}\Delta m = 1.642354 \times 10^{-28} \, \text{kg} \\ \Delta E = c^2 \Delta m \\ \Delta E = (3 \times 10^8)^2 \times (1.642354 \times 10^{-28}) \\ \Delta E = 1.478 \times 10^{-11} \, \text{J} \end{gather}[/math]
Converting to electron volts
[math]\begin{gather}\Delta E = \frac{1.478 \times 10^{-11}}{1.6 \times 10^{-19}} \\ \Delta E = 9.226 \times 10^7 \, \text{eV} \\ \Delta E = 92.3 \, \text{MeV} \end{gather}[/math]
Alternatively, working in atomic mass units
[math]\begin{gather}
\Delta m = 0.098937 \, u \\
\Delta E = 931.5 \times 0.09893 \\
\Delta E = 92.2 \, \text{MeV}
\end{gather}[/math]
135) Understand the processes of nuclear fusion and fission with reference to the binding energy per nucleon curve
⇒ Binding Energy per Nucleon:
To work this out we need to know both the binding energy of a nucleus, and the number of nucleons within it. This gives us the binding energy per nucleon in a nucleus in MeV, and from this we can determine how strongly different nuclei are held together.
Figure 1 Graph of binding energy per nucleon against mass number, A.
Drawing a graph of binding energy per nucleon against mass number for the nuclei gives us a useful way of comparing how tightly different nuclides are bound together.
The isotope with the highest binding energy per nucleon is iron-56 at 8.8 MeV per nucleon. Any nuclear reaction which increases the binding energy per nucleon will give out energy.
The graph shows us that small nuclides can combine together to make larger nuclei (up to Fe-56) with a greater binding energy per nucleon. This process is called nuclear fusion.
Similarly, larger nuclei can break up into smaller pieces which have a greater binding energy per nucleon than the original nucleus. Reactions like this are called nuclear fission. Both of these types of nuclear reaction will give out energy, and could be used as power sources.
136) Understand the mechanism of nuclear fusion and the need for very high densities of matter and very high temperatures to bring about and maintain nuclear fusion
⇒ Nuclear Fusion:
If we take some light nuclei and force them to join together, the mass of the new heavier nucleus will be less than the mass of the constituent parts, as some mass is converted into energy.
However, not all of this energy is used as binding energy for the new larger nucleus, so energy will be released from this reaction.
The binding energy per nucleon afterwards is higher than at the start. This is the process of nuclear fusion and is what provides the energy to make stars shine (figure 2).
Figure 2 The proton-proton nuclear fusion reaction, typical in small cool stars such as our Sun, where the core temperature is about [math]15 × 10^6[/math] K.
These reactions which occur in enormous numbers provide the energy which causes the Sun to emit heat and light in all directions in great quantities.
Converting hydrogen into helium with the release of energy could be an excellent way of supplying the planet’s energy needs.
The seas are full of hydrogen in water molecules, and the helium produced would be an inert gas which could simply be allowed to go up into the upper atmosphere.
However, scientists have not yet successfully maintained a controlled nuclear fusion reaction. The problem lies in forcing two positively charged, mutually repelling, protons to fuse together.
The kinetic energy which they need to collide forcefully enough to overcome this electrostatic repulsion requires temperatures of many million kelvin.
Moreover, to ensure enough colliding protons for the reaction to be sustained requires a very high density of them. Comparing the energy output with nuclear fission below, 235 grammes of hydrogen undergoing nuclear fusion would produce an energy output of [math]1.40 × 10^{14} J[/math].
⇒ Nuclear Fission:
Nuclear fusion is not yet an option for electricity generation, as we have not been able to create the high densities and temperatures needed to sustain a fusion reaction.
However, another process which releases binding energy from nuclei is called nuclear fission.
In this process a large nucleus breaks up into two smaller nuclei, with the release of some neutrons and energy (figure 3).
Figure 3 Nuclear fission of uranium-235.
Fission reactions can be started when the nucleus absorbs another particle which makes it unstable. Uranium-235 can absorb slow-moving neutrons to become the unstable isotope U-236.
Example:
Although there are many different possible products from nuclear fission reactions, the amount of energy released in each one is about the same at about 200 MeV. Calculate the energy released in the following fission reaction:
[math]U^{235} + n → Ba^{141} + Kr^{92} + 3n + energy[/math]
Solution:
Mass of U-235 = 235.0439 u
Mass of Be-141 = 140.9144 u
Mass of Kr-91.9262 u
Mass difference ∆m
[math]\begin{gather}
\Delta m = 235.0439 + 1.008665 – 140.9144 – 91.9262 – 3(1.008665) \\
\Delta m = 0.1860 \, u \\
\text{Energy released } E = 0.1860 \times 931.5 \\
E = 173 \, \text{MeV} \\
E = 2.77 \times 10^{-11} \, \text{J}
\end{gather}[/math]
This is the energy per fission. If one mole of these were to occur, using 235 g of U-235, then the total energy produced would be:
[math]\begin{gather}
E = 6.02 \times 10^{23} \times 2.77 \times 10^{-11} \\
E = 1.67 \times 10^{13} \, \text{J}
\end{gather}[/math]
However, U-235 is a small proportion (≈0.7%) of all the uranium found, and thus a larger amount of uranium fuel would be needed in order to provide enough U-235 atoms to produce this much energy.
137) Understand that there is background radiation and how to take appropriate account of it in calculations
Human beings can survive small doses of nuclear radiation. This has been important in our evolution, as the natural environment has low levels of radiation from natural sources. This is called background radiation.
The level varies from place to place around the world. For example, in the UK it averages to less than one radioactive particle every two seconds in any given place (fig A).
If we measure background radiation using a Geiger-Müller (G-M) tube, the number of counts per second usually ranges from 0.2-0.5 depending on exact location.
Radiation levels are often reported in counts per second, and this unit is called the becquerel (Bq). Henri Becquerel was the French physicist who discovered spontaneous radiation in 1896.
Figure 4 The environment continually exposes us to low levels of nuclear radiation.
The precise amount of radioactive radiation that a person is exposed to from their surroundings depends on their location and duration of stay. This is due to several environmental
The local background radiation level is influenced by several variables. Naturally occurring radioactive gases in the atmosphere, especially radon, account for around half of the background radiation on Earth.
Radon is more prevalent in some regions of the world than others because it is created during the breakdown of uranium ore, which is found in some rocks, particularly granite.
To eliminate excess radon gas from the home atmosphere, several homes in Cornwall, UK, are equipped with radon detectors and specialized ventilation systems.
138) Understand the relationships between the nature, penetration, ionizing ability and range in different materials of nuclear radiations (alpha, beta and gamma)
139) Be able to write and interpret nuclear equations given the relevant particle symbols
Many nuclei are slightly unstable and there is a small probability that, each second, they may decay.
This means that a nucleon may change from one type to another, or the composition or energy state of the nucleus as a whole may change.
When a nuclear decay occurs, the radiation particle emitted will leave the nucleus carrying kinetic energy.
As the particle travels, it will ionize particles in its path, losing some of that kinetic energy at each ionization.
When all the kinetic energy is transferred, the radiation particle stops and is absorbed by the substance that it is in at that moment.
The three main types of nuclear radiation are called alpha (α), beta (β) and gamma (γ) radiation.
Each one comes about through a different process within the nucleus, each one is composed of different particles, and each one has different properties.
⇒ Alpha decay:
Alpha particles are composed of two protons and two neutrons, the same as a helium nucleus. This is a relatively large particle with a significant positive charge ([math][/math]), so it is highly ionizing.
An alpha particle moving through air typically causes 10 000 ionizations per milli-metre. As it ionizes so much, it quickly loses its kinetic energy and is easily absorbed.
A few centi-metres travel in air is enough to absorb an alpha particle, and they are also blocked by paper or skin.
[math]{}^{241}_{95}\mathrm{Am} \rightarrow {}^{237}_{93}\mathrm{Np} + {}^{4}_{2}\alpha[/math]
Figure 5 Alpha particle emission.
⇒ Beta Decay:
A beta particle is an electron emitted at high speed from the nucleus when a neutron decays into a proton. With its single negative charge and much smaller size, the beta particle is much less ionizing than an alpha particle, and thus can penetrate much further. Several meters of air, or a thin sheet of aluminum, are needed to absorb beta particles.
[math]{}^{41}_{6}\mathrm{C} \rightarrow {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}\beta + \bar{\nu}_{0}^{0}[/math]
Figure 6 Beta-minus particle emission.
⇒ Gamma Decay:
Gamma rays are high energy, high frequency, electromagnetic radiation. These photons have no charge and no mass and so will rarely interact with particles in their path, which means they are the least ionizing nuclear radiation.
They are never completely absorbed, although their energy can be significantly reduced by several centi-meters of lead or several meters of concrete. If the energy of gamma rays is reduced to a safe level, we often describe them as being absorbed.
Example gamma decay equation:
[math]{}^{60}_{28}\mathrm{Ni}^m \rightarrow {}^{60}_{28}\mathrm{Ni} + {}^{0}_{0}\gamma[/math]
Figure 7 Gamma ray emission. The image shows only the gamma emission from an excited nickel nucleus, Ni-m a moment after a beta particle was emitted from a cobalt-60 nucleus.
Gamma ray emission does not change the structure of a nucleus emitting it, but carries away energy.
So, the nucleus must drop energy level in order to emit the gamma ray. This most commonly happens immediately after alpha or beta decay, which leaves the nucleus in an excited state.
140) CORE PRACTICAL 15: Investigate the absorption of gamma radiation by lead
⇒ Investigating lead absorption of gamma radiation:
Using the same experimental set-up as shown in fig B, we can find out how much lead is needed to reduce the intensity of gamma radiation by a particular amount.
Using a radioactive source that emits alpha and gamma only, we can select to study the gamma radiation by placing a sheet of paper in front of the source to absorb all the alpha particles.
This will have no measurable effect on the gamma emissions. Keeping the distance from the source to the G-M tube constant, we can place different thicknesses of lead plate in the gap between them.
Figure 8 Investigating nuclear radiations.
For each thickness, record the count over a fixed time period and determine the corrected count rate. There is a mathematical relationship between the corrected count rate (which represents intensity of the radiation, I) and the thickness, x, of lead. Work out what graph to draw to prove that
[math]I ∝ e^{-μx}[/math]
the penetrating power of alpha, beta and gamma radiation using a Geiger-Müller tube to detect them. Between the source and the G-M tube, place absorber sheets that progressively increase in density, and measure the average count rate.
This investigation is often simulated using computer software. This removes all risk of exposure to radiological hazards.
141) Understand the spontaneous and random nature of nuclear decay
142) Be able to determine the half-lives of radioactive isotopes graphically and be able to use the equations for radioactive decay
[math]\text{activity } A=\lambda N, \, \frac{dN}{dt}=-\lambda N, \, \lambda=\frac{\ln 2}{t_{1/2}}, \, N=N_0 e^{-\lambda t} \text{ and } A=A_0 e^{-\lambda t}[/math]
Derive and use the corresponding log equations.
⇒ Probability and Decay:
Radioactive decay is a spontaneous and random process and so any radioactive nucleus may decay at any moment. For each second that it exists, there is a particular probability that the nucleus will decay.
This probability is called the decay constant, X. However, just like guessing which number will come up next in a lottery, it is not possible to predict when any given nucleus will decay.
The chance that a particular nucleus will decay is not affected by factors outside the nucleus, such as temperature or pressure, or by the behavior of neighboring nuclei – each nucleus decays entirely independently.
If we have a large sample of the nuclei, the probability of decay will determine the fraction of these nuclei that will decay each second. Naturally, if the sample is larger, then the number that decay in a second will be greater. So, the number decaying per second, called the activity
[math]A \text{or} (\frac{dN}{dt})[/math] is proportional to the number of nuclei in the sample, N. Mathematically, this is expressed as:
[math]A = -\lambda N \\
\frac{dN}{dt} = -\lambda N[/math]
The minus sign in this formula occurs because the number of nuclei in the sample, N, decreases with time. In practice we ignore the negative sign when using the formula.
The formula for the rate of decay of nuclei in a sample is a differential equation. We have previously met this type of equation when studying the discharge of a capacitor.
The equation
[math]\frac{dN}{dt} = -\lambda N[/math]
can be solved to give a formula for the number of nuclei remaining in a sample, N, after a certain time, t.
[math]N = N_0 e^{-λt}[/math]
Where [math]N_o[/math] is the initial number of nuclei within a sample.
Half- Life:
the activity of a radioactive sample decreases over time as the radioactive nuclei decay, leaving fewer radioactive nuclei available to decay.
While the activity of a sample depends on the number of nuclei present, the rate at which the activity decreases depend only on the particular isotope. A measure of this rate of decrease of activity is the half-life [math]t_{1/2}[/math]
This is the time taken for half of the nuclei of that nuclide within a sample to decay.
Mathematically, the half-life can be found by putting
[math]\begin{gather}
N = \frac{1}{2} N_0 \\
\text{Into the decay equation} \\
N = N_0 e^{-\lambda t} \\
\frac{1}{2} N_0 = N_0 e^{-\lambda t_{1/2}} \\
\frac{1}{2} = e^{-\lambda t_{1/2}} \\
\ln \frac{1}{2} = -\lambda t_{1/2} \\
t_{1/2} = \frac{\ln 2}{\lambda}
\end{gather}[/math]
Rearranging this also gives us
[math]\lambda = \frac{\ln 2}{t_{1/2}}[/math]
Example:
The decay constant for carbon-14 is [math]λ = 3.84 × 10^{-12} s^{-1}[/math], What is the activity of a sample of 100 billion atoms of carbon-14?
[math]\begin{gather}
\frac{dN}{dt} = -\lambda N \\
\frac{dN}{dt} = -(3.84 \times 10^{-12})(100 \times 10^{9}) \\
\frac{dN}{dt} = 0.384 \, \text{Bq}
\end{gather}[/math]
Example:
What is the half-life of carbon-14?
[math]\begin{gather}
\lambda = 3.84 \times 10^{-12} \, \text{s}^{-1} \\
t_{1/2} = \frac{\ln 2}{3.84 \times 10^{-12}} \\
t_{1/2} = 1.81 \times 10^{11} \, \text{s} \\
t_{1/2} = 5720 \, \text{years}
\end{gather}[/math]

For video lectures subscribe to our YouTube channel

Request a lesson

Pearson Edexcel Physics

Unit 5: Thermodynamics, Radiation, Oscillations and Cosmology

5.4 Nuclear Decay

Pearson Edexcel Physics

Unit 5: Thermodynamics, Radiation, Oscillations and Cosmology

5.4 Nuclear Decay

133) Understand the concept of nuclear binding energy and be able to use the equation [math]ΔE = c^2 Δm[/math] in calculations of nuclear mass (including mass deficit) and energy

134) Use the atomic mass unit (u) to express small masses and convert between this and SI units

135) Understand the processes of nuclear fusion and fission with reference to the binding energy per nucleon curve

136) Understand the mechanism of nuclear fusion and the need for very high densities of matter and very high temperatures to bring about and maintain nuclear fusion

137) Understand that there is background radiation and how to take appropriate account of it in calculations

138) Understand the relationships between the nature, penetration, ionizing ability and range in different materials of nuclear radiations (alpha, beta and gamma)

139) Be able to write and interpret nuclear equations given the relevant particle symbols

140) CORE PRACTICAL 15: Investigate the absorption of gamma radiation by lead

141) Understand the spontaneous and random nature of nuclear decay

142) Be able to determine the half-lives of radioactive isotopes graphically and be able to use the equations for radioactive decay

[math]\text{activity } A=\lambda N, \, \frac{dN}{dt}=-\lambda N, \, \lambda=\frac{\ln 2}{t_{1/2}}, \, N=N_0 e^{-\lambda t} \text{ and } A=A_0 e^{-\lambda t}[/math]

Derive and use the corresponding log equations.

TIPS FOR STUDENTS TO PREPARE FOR PHYSICS FOR AQA A LEVEL

2023/08/03 at 4:00 pm

A level physics past papers with answers pdf

2023/09/20 at 4:25 pm

Practicing Physics past papers for AQA A LEVEL

2023/06/13 at 8:11 am

Top 10 Maths & Physics Tutors Near You

2023/01/08 at 8:25 pm

Subscribe to our news Letter

We've added you to our list. Further Please Check your Email account for verification