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HSC Physics for NSW Study Guide Series
A more detailed explanation can be found by searching YouTube for Chadwick and the neutron high
school physics explained . Open
Watch this through once to get the basics and then watch again stopping when necessary to answer the
following questions.
Questions:
Why is the atomic number smaller than the atomic mass?
Describe, with the help of a labelled diagram, the experiment carried out by Bothe and Becker.

What modification did Curie and Joliet make to the Bothe and Becker experiment?

Chadwick measured the velocity of the protons ejected from the paraffin. Explain the significance of this.

Chadwick applied the conservation of momentum and kinetic energy to his measurements. What conclusion
did he draw?

46 Module 8 From the Universe to the Atom

Summary
In 1920, Rutherford suggested that a proton inside the nucleus might have an electron attached so close to
it as to form a neutral particle for which he suggested the name "neutron."
The difficulties in the search for neutrons were:

no naturally occurring materials emitted neutrons
methods of detecting particles depend on their electric charge which the neutron does not have.

In 1930 it was found that boron and beryllium emitted radiation when hit by an alpha particle.

beryllium or paraffin block
boron

radioactive source an
alpha emitter

detector

protons

alpha particles

unknown radiation
which we now know
are neutrons

The radiation from the beryllium or boron was:

uncharged similar to gamma radiation

more penetrating than gamma radiation and had a very high energy. This energy was higher than any
gamma ray discovered up until then. Therefore, the possibility was that this radiation was not gamma
radiation.

Further studies showed that:
the radiation was absorbed by paraffin, a material which contains a lot of hydrogen
the paraffin then ejected protons.

Conservation of kinetic energy and momentum
Chadwick performed two experiments in which he had the unknown particles emitted from the beryllium
hit:

paraffin

nitrogen.

By measuring the speed of the protons emitted from the paraffin and the speed of the nitrogen atoms
Chadwick, using the conservation of kinetic energy and momentum, was able to calculate the properties of
the unknown radiation.

Chadwick discovered that the unknown radiation must have a similar mass as a proton. As the unknown
radiation was uncharged, this meant that the radiation emitted from the beryllium must have been the
particle which Rutherford had named as a neutron.

The reaction with beryllium was: 4 + 42 → 162 + 01

Module 8 From the Universe to the Atom 47

HSC Physics for NSW Study Guide Series
Properties of neutrons

Neutrons easily penetrate material because they have no charge. They are not repelled by the positive
nuclei.
The discovery of neutrons led to a new model of the nucleus consisting of protons and neutrons with
no electrons in the nucleus.
Neutrons easily explain the existence of isotopes. Isotopes are atoms with the same number of protons
but different numbers of neutrons in the nucleus. Extra neutrons in the nucleus do not alter the chemical
properties of the element as the electron arrangement is the same.
Question sheet 8.5
1. Describe, with the aid of a labelled diagram, Chadwick s experiment on the discovery of the neutron.

2. Rutherford predicted that the neutron existed and that it is equivalent to a proton and an electron.
Explain why:

(a) the neutron has no charge?

(b) the neutron has a mass slightly more than the proton?

3. Distinguish how the isotopes of elements are: (a) similar to one another, and (b) different from one
another.

4. In an experiment two parallel metal plates, 3.2 x 10-2 m apart, are connected to a 600 V power
supply. The electric field between the plates just supports the weight of an oil drop of mass 1.8 x
10-14 kg, which has a charge due to a few excess electrons.

(a) Calculate the charge on the oil drop.

(b) What is the most likely number of excess electrons acquired by the oil drop?

48 Module 8 From the Universe to the Atom

Inquiry question: How is it known that atoms are made up of protons, neutrons and electrons?

Now that you have covered this section you should be better able to answer the inquiry question with your new
understanding of the structure of the atom.

Summary
Write a summary of the structure of the atom, review your summary with others in the class, update as
necessary.

.
.
.
.
.
.

.
.
.
.

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HSC Physics for NSW Study Guide Series

Quantum Mechanical Nature of the Atom

Students:
assess the limitations of the Rutherford and Bohr atomic models

investigate the line emission spectra of elements to examine the Balmer series in hydrogen (ACSPH138)

relate qualitatively and quantitatively the quantised energy levels of the hydrogen atom and the law of

conservation of energy to the line emission spectrum of hydrogen using:

- = ℎ

- = ℎ


- 1 = 1 − 1 (ACSPH136)
2 2


investigate de Broglie s matter waves, and the experimental evidence that developed the following formula:

- = ℎ (ACSPH140)


analyse the contribution of Schrödinger to the current model of the atom

NSW Physics Stage 6 syllabus © NSW Education Standards Authority for and on behalf of the Crown in right of the State of New South
Wales, 2017.

Inquiry question: How is it known that classical physics cannot explain the properties of the atom?

With all inquiry questions consult with class members, come to a consensus and then summarise your answer in the
space provided.

50 Module 8 From the Universe to the Atom

Limitations of the Rutherford atomic model
As we saw previously Rutherford s model of the atom has the electron orbiting a positive nucleus. This is
a problem as it disagrees with classical physics which says that an electron orbiting the nucleus should
radiate electromagnetic radiation continually and spiral into the nucleus.
Another limitation with the Rutherford model is that it could not explain the spectra of gases, for example,
the lines in the emission spectrum of hydrogen.
The Bohr atomic model
The syllabus requires student to assess the limitations of the Bohr atomic model. However, before we can
do this, we need to review the model.
Task
Search YouTube for The Bohr Atom Bozeman Science Openv and then answer the following questions.
Questions:
What did Bohr realise must happen with moving charged particles?

What kind of spectra did Bohr think should be produced with the Rutherford model?

Which element does the Bohr model of the atom work with?
How does an electron move between energy levels in a Bohr atom?

Discuss the ladder analogy that is used in this video.

Complete the following table for the Balmer series. The first transition has been done for you.
Note: you do not need to calculate instead copy the values from the diagram in the video.

Electron movement wavelength (nm)
three to two 656

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HSC Physics for NSW Study Guide Series

Summary
The Bohr atomic model
In order to explain the emission spectra of hydrogen, Bohr suggested that electrons in atoms follow these
rules:

1. Electrons orbit the nucleus in what are described as stationary or stable states without emitting
electromagnetic radiation. As long as the electron is in one of these energy orbits it will not lose or
absorb any energy.

2. Each circular orbit has a different radius about the nucleus.

3. Electrons have only certain energies corresponding to particular distances from the nucleus. The energy
of the orbits can be compared to the steps on a ladder: electrons can only be on the steps of the ladder
and not in between steps.

4. Any change in the orbit of the electron must be accompanied by a movement from one stationary state
to another.

5. Movement between stationary states emits or absorbs electromagnetic radiation.

6. When an electron moves between stationary states, it releases or absorbs a photon. The photon's energy
is quantised and given by E = hf where E is the energy difference between the stationary states.

With these rules and using a mixture of classical and quantum physics Bohr was able to develop a
mathematical model which could predict the wavelengths present in the hydrogen spectrum.

The Bohr model gave the same relationship as Balmer had produced empirically for the hydrogen spectrum.

Limitations of Bohr s atomic model
Bohr s model explained the behaviour of electrons in the hydrogen atom and therefore the spectrum of the
hydrogen atom. He chose hydrogen because it was the simplest atom but could not manage a theory which
explained:

the spectra of larger atoms
the splitting of the spectral lines when there is a magnetic field (the Zeeman effect)
the relative intensity (the difference in brightness) of the lines in the hydrogen spectra
the existence of hyperfine spectral lines. These are spectral lines which consist of a series of very fine,
closely spaced lines.

The spectra of larger atoms
While the energy of electron movements for all elements is quantised, Bohr's explanation and prediction of
wavelengths works only for hydrogen. This may be because of interactions between electrons, the shielding
electrons get from inner levels of electrons, non-circular orbits or stronger attractions of electrons to more
positive nuclei - whatever the reason, the Bohr model works only for hydrogen.

The Zeeman Effect
In 1896, Zeeman observed the spectrum of sodium in a magnetic field. Some lines were split into three.
This is called the Zeeman effect. The Zeeman effect could not be explained by the Bohr model.

The relative intensity of spectral lines
Lines in the spectrum of hydrogen have varying intensity and thickness. Some are bright, some broad, some
sharp, others thin, and some dull. Bohr's model cannot explain these differences.

Hyperfine spectral lines
Hyperfine spectral lines are the spectral lines formed from the splitting of broader spectral lines due to a
property called nuclear "spin". Electrons are said to spin up or spin down . These hyperfine lines could
not be explained by classical physics or the Bohr model.

52 Module 8 From the Universe to the Atom

Practical investigation examining the Balmer series in Hydrogen

Your school may have a hydrogen tube whose light you can examine with a spectroscope. As an alternative
or to prepare for this investigation search YouTube for Emission spectrum of Hydrogen Khan Academy .
Open
Summarise what you learnt from this video and any school investigation. Make sure you sketch the
hydrogen spectrum you observed including the colours.
Safety: review the risk assessment on spectral tubes from module 7 - the nature of light.

Another common spectral tube in schools is sodium. Observe this light though a spectroscope and describe
your observations below.

Where else do we observe a sodium light source?

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HSC Physics for NSW Study Guide Series

The emission spectrum of hydrogen
This consists of a series of lines which correspond to specific colours and wavelengths as shown below.
You should be able to observe four bright coloured lines.

420
434
486
656

wavelength in nm

The Balmer series
The hydrogen spectrum consists of a series of lines which correspond to specific colours and wavelengths
as shown above. These are described as discrete, single line, spectra.
Balmer looked at these lines and eventually worked out empirically that all four wavelengths fit into the
equation:

1 11
= 2 − 2
where:

R is the Rydberg constant for Hydrogen R = 1.097 x 107 m-1
nf = integer which represents the final energy level of the atom
ni = integer which represents the initial energy level of the atom

We use this relationship to calculate the wavelength of the hydrogen emission lines.

Note: Bohr obtained the same relationship but used a mathematical theory which is beyond the scope of
this course.

Sample problem 8.4
An electron falls from n = 3 to the n = 2 level in a hydrogen atom. Calculate the wavelength of the emitted
light.

Solution:

1 = 1 − 1 = 1.097 x 107 (212 − 312).
2 2


This gives a wavelength of 656 nm.

Question sheet 8.6
Take RH the Rydberg constant = 1.097 x 107 m-1

1. Outline the limitations of the Rutherford and Bohr atomic models.

54 Module 8 From the Universe to the Atom

2. Outline with the help of a labelled diagram an investigation to observe the Balmer series.

3. The Bohr model was a stepping stone towards a good model of the electronic structure of the atom.
Outline the rules that Bohr put forward to explain the hydrogen spectra.

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HSC Physics for NSW Study Guide Series

Quantised energy levels of the hydrogen atom

When electrons move between energy levels then − = ℎ = ℎ

where h = Planck s constant = 6.626 x 10-34 J s

Ei = the initial excited state energy level of the electron in Joules

Ef = the final lower energy level of the electron in Joules

f = frequency of the light c = speed of light = wavelength of light

Energy levels can be represented by diagrams as shown below. The ground state of the electron can be
labelled as 0 eV, hence, excited energy levels have a positive value. Alternatively, the ground state may be
given a negative value and when the electron has been removed, that is ionised, labelled as 0 eV.

The energy levels in hydrogen are quantised and are given by the following relationships:

For negative values = − 0 where E0 = 13.60 eV and n is an integer.
2
13.6
For positive values E = 13.6 - 2 .

This gives the following diagrams:

with negative energy in eV with positive energy in eV

0 n= 13.60
- 0.54 n=5 13.06

- 0.85 n=4 12.75

- 1.51 n=3 12.09

- 3.40 n=2 10.20

- 13.60 n=1 0

Sample problem 8.5
An electron in a hydrogen atom moves from level 5 to 2. Calculate the wavelength of the light released.

Solution:
Note in this case energy needs to be converted from eV to Joules.

Using negative energy 0.54 = 2.86 x 1.602 x 10-19 J = 4.58 x 10-19 J

− = (- 0.54 - - 3.40) eV = 3.40
4.58 x 10-19 = h x f = 6.626 x 10-34 x f
f = 6.91 x 1014 Hz

v=f
= 3 x 108/6.91 x 1014 = 434 nm

Using positive energy

− = (13.06 10.20) eV = 2.86 eV
This gives the same value, 434 nm as above.

Using the relationship

1 = 1 − 1 = 1.097 x 107 (212 − 12)
2 2


This gives the wavelength as 434 nm, the same as above.

56 Module 8 From the Universe to the Atom

Question sheet 8.7
1. The diagram below shows some of the energy levels of hydrogen. In this case the energy levels are

given as positive values. The transition between energy levels is represented by an arrow.

with positive energy in eV

n= 13.60
n=5 13.06
n=4
12.75
n=3
12.09
n=2
10.20

n=1 0

(a) Explain why the light produced in the transition from n = 4 to n = 3 has a longer wavelength than
the light produced in the transition from n = 4 to n = 2.

(b) Calculate the wavelength of the photon emitted when a hydrogen atom undergoes the transition
from n = 5 to n = 1.

(c) What region (visible, infrared, or ultraviolet) is the electromagnetic spectrum of this photon in part
(b).

2. A tungsten filament light globe does not produce bright line spectra. Explain why this is the case.

3. An electron in hydrogen drops from energy level, n = 4 of binding energy 1.36 x 10-19 J, to energy level
n = 1 of binding energy 21.76 x 10-19 J. Calculate the energy emitted by the electron in making this

transition. Note in this case binding energy means that the values are negative.

Module 8 From the Universe to the Atom 57

HSC Physics for NSW Study Guide Series energy in eV n=5
n=4
4. An electron falls from energy level n = 4 to n = 1 - 0.38 n=3
as shown in the diagram opposite. Calculate the - 0.55
frequency and wavelength of the light emitted. n=2
Note this is not a hydrogen atom.

- 0.85

- 1.51

- 3.41 n=1

5. The diagram shows the energy levels for a hydrogen atom. If a photon of UV light of wavelength 97.4
nm is absorbed by the hydrogen atom to which level will an electron in energy level n = 1 move? Show
your working.

Energy in joules

n= 2.20 x 10-18
n=5 2.08 x 10-18

n=4 2.04 x 10-18

n=3 1.94 x 10-18

n=2 1.63 x 10-18

n=1 0

6. (a) A red line in the hydrogen spectrum is produced when electrons move from the 3rd energy level to
the 2nd energy level. Calculate the wavelength and photon energy in both Joules and eV of the red
light.

(b) Calculate the wavelength and energy emitted when an electron in a hydrogen atom moves from the
ni = 2 level to the nf = 1 level. Note this wavelength is in the UV region and is not part of the Balmer
series.

58 Module 8 From the Universe to the Atom

de Broglie and matter waves
de Broglie proposed that as light and other electromagnetic radiations have both wave and particle
properties then particles that are moving may also have wave properties.

de Broglie said that a particle of mass m and velocity v has a wavelength given by: = ℎ or = ℎ



where:
h = Planck s constant = 6.626 x 10-34 J s
= wavelength of the particle
m = mass of the particle
v = velocity of the particle
p = momentum of the particle

Davisson and Germer s experiment
In 1927, Davisson and Germer conducted an experiment which experimentally verified de Broglie s
hypothesis by showing the diffraction of electrons. The wavelength of the electron was in agreement with
what de Broglie had predicted.

Task
Search YouTube for A level Physics the de Broglie wavelength and wave particle duality and then answer
the following questions. Open

Complete the following: if a wave can behave as a particle then a particle can behave as a ..

Draw a labelled diagram of the experimental setup shown at 2 minutes 30 seconds into the video.

Outline the observations made in the experiment. 59
What was the significance of the observed diffraction pattern of the electrons?

Module 8 From the Universe to the Atom

HSC Physics for NSW Study Guide Series

Summary
Davisson and Germer s experiment consisted of an electron beam fired from an electron gun at 90 degrees
to the surface of a Nickel crystal. The experiment was placed in a vacuum to avoid collisions between
electrons and air molecules.
During the experiment, air accidentally entered the tube producing an oxide film on the nickel surface. To
remove the oxide Davisson and Germer heated the nickel to a high temperature not realising that this would
turn the nickel into a large single crystal. This nickel crystal was essential to the success of the experiment.

Observations
Davisson and Germer observed that electrons were diffracted showing constructive and destructive

interference which is a wave property.
They calculated the wavelength of the electron from the accelerating voltage V by using the following
relationships:
Kinetic energy = W or ½ mv2 = qV

Rearranging this gives = 2 and as = ℎ this gives = ℎ or = ℎ
2 2


The wavelength from this equation agreed with their observations. This confirmed that the electron has a
wave nature.

de Broglie and the hydrogen atom
The suggestion made by de Broglie that electrons can behave as waves explained why the energy of
electrons is quantised. If we assume that electrons orbit the nucleus with a standing wave pattern then it
requires an integer number of wavelengths to fit in the circumference of an orbit. You may need to revise
the section on standing waves in module 3 -Waves and Thermodynamics.
We can investigate this as follow with the equipment shown below. A loop of wire is attached to a vibration
generator connected to an amplifier and signal generator. At specific frequencies standing waves are
observed in the loop. This is similar to the standing wave pattern we have of electrons orbiting a nucleus.
This occurs at fixed frequencies, that is the energy is quantised which was one of Bohr s suggestions (also
called postulates). The wave nature of electrons explained the reason for this postulate.
Your school may have this equipment or as an alternative search YouTube for the following: Standing

Wave Demo: Vibrating Hoop Physics Demos . Open

Summarise your observations below:

standing waves

wire loop

vibration
generator

Note: the electron as a standing wave requires whole number of wavelengths in the circumference. That is
the circumference = n where n = 1, 2, 3, etc.

60 Module 8 From the Universe to the Atom

Question sheet 8.8
1. Calculate the wavelength of a car of mass 1500 kg travelling at 30 m s-1.

2. Calculate the wavelength of an electron of mass 9.1 x 10-31 kg travelling at 2.8 x 107 m s-1.

3. Discuss the results of questions 1 and 2.

4. Outline Davisson and Germer s confirmation of the wave nature of electrons. Include a labelled diagram
in your answer.

5. What is the speed of an electron with a de Broglie wavelength of 0.260 nm?

6. What is the de Broglie wavelength of an electron with 13.6 eV of kinetic energy?

7. Neutrons may be used to study the atomic structure of matter. Diffraction effects are noticeable when
the de Broglie wavelength of the neutrons is comparable to the spacing between the atoms. This spacing
is typically 2.6 x 10-10 m.
(a) Suggest why using neutrons may be preferable to using electrons when investigating matter.

(b) Calculate the speed of a neutron having a de Broglie wavelength of 2.6 x 10-10 m. The mass of a
neutron is 1.7 x 10-27 kg.

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HSC Physics for NSW Study Guide Series
Research task
The syllabus asks students to analyse the contribution of Schrödinger to the current model of the atom.
Search YouTube for the following videos to help you.

Quantum Mechanics and the Schrodinger Equation by Professor Dave Explains Open and What is The
Schrödinger Equation, Exactly? by Up and Atom . Open
Make notes on a sheet of paper then get together in small groups to compare your analysis.
Once you have reviewed and modified your work then summarise below. Try and keep it simple only put
in material which you understand.

62 Module 8 From the Universe to the Atom

Inquiry question: How is it known that classical physics cannot explain the properties of the atom?

Now that you have covered this section you should be better able to answer the inquiry question with your new
understanding of classical physics and why it cannot explain the properties of the atom.

Summary
Write a summary of the quantum mechanical nature of the atom, review your summary with others in the
class, update as necessary.

.
.
.
.
.
.

.
.
.
.

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HSC Physics for NSW Study Guide Series

Properties of the Nucleus

Students:
analyse the spontaneous decay of unstable nuclei, and the properties of the alpha, beta and gamma radiation
emitted (ACSPH028, ACSPH030)
examine the model of half-life in radioactive decay and make quantitative predictions about the activity or
amount of a radioactive sample using the following relationships:

- = −
- = 2

1/2

where = number of particles at time , 0 = number of particles present at = 0, = decay constant,
1/2 = time for half the radioactive amount to decay (ACSPH029)

model and explain the process of nuclear fission, including the concepts of controlled and uncontrolled chain
reactions, and account for the release of energy in the process (ACSPH033, ACSPH034)
analyse relationships that represent conservation of mass-energy in spontaneous and artificial nuclear
transmutations, including alpha decay, beta decay, nuclear fission and nuclear fusion (ACSPH032)
account for the release of energy in the process of nuclear fusion (ACSPH035, ACSPH036)
predict quantitatively the energy released in nuclear decays or transmutations, including nuclear fission and
nuclear fusion, by applying: (ACSPH031, ACSPH035, ACSPH036)
- the law of conservation of energy
- mass defect
- binding energy
- Einstein s mass energy equivalence relationship ( = 2)

NSW Physics Stage 6 syllabus © NSW Education Standards Authority for and on behalf of the Crown in right of the State of New South
Wales, 2017.

Inquiry question: How can the energy of the atomic nucleus be harnessed?

With all inquiry questions consult with class members, come to a consensus and then summarise your answer in the
space provided.

64 Module 8 From the Universe to the Atom

The spontaneous decay of unstable nuclei

Radioactivity is the spontaneous disintegration, that is decay, of an atom as an unstable nucleus tries to
reach a more stable nuclear configuration. The process of radioactive decay occurs in general by three
methods:

a nucleus that changes one of its neutrons into a proton with the simultaneous emission of an electron -
this is beta decay

by emitting a helium nucleus this is alpha decay

by spontaneous fission, that is splitting into two fragments.

After radioactive decay the new atom is often still radioactive. This means that the atom will decay again
as it attempts to reach a stable state. It often does this by emitting gamma rays.

Before using any radioactive sources, a risk assessment should be carried out.

Risk Assessment: Radioactive Sources

Activity Using radioactive sources.

Hazard Contamination with radioactive materials. Exposure to radioactivity.

Hazard information Radioactivity can cause cell damage.

Summary of activity
Measuring the penetrating power of radiation through air, paper, aluminium and lead.

Risk identification
Long term exposure to radiation may cause cancer.

Discuss with your class where you think the ticks should go.

Likelihood very likely likely unlikely very unlikely
(tick box)

Consequence extreme major moderate minor
(tick box)

Control
Sources should be handled wearing gloves and held using tongs to increase distance. Remember the
inverse square law. Wash hands after use and dispose of gloves in bin.

Write a conclusion about the risk in the space below.

Conclusion
about risks

Student involved in assessment:

Name of Assessor: .

Signature: . Date:
(of assessor)

Note the assessor is your teacher who has discussed with you and approved the risk assessment.

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Natural radioactivity
Some elements decay naturally by undergoing alpha, beta or gamma decay. Examples of these follow:

Alpha decay

232 2304 ℎ + 24 Uranium 238 decays into Thorium 234 and an alpha particle

Note: the atomic and mass numbers must balance on both sides of the equation.

Beta decay

2340 ℎ → 2341 + −01 Thorium 234 decays into Protactinium 234 and a beta particle

Note: the atomic number increases by one during beta decay as a neutron decays into a proton and electron.
That is: 10 → 11 + −01 . We look at this transformation later in terms of particles called quarks.

Gamma decay
With gamma decay there is no change in mass or atomic number. It is observed that gamma rays do not
appear on their own. They are only emitted with a beta or alpha particle. The atom emits a gamma ray to
release the excess energy it has after emitting an alpha or beta particle.

These reactions all rely on the atom going to a more stable state, where energy is released so the mass of
the products on the right-hand side is smaller than the mass of the reactants on the left-hand side.

Decay series
Radioactive decay does not normally involve one atom only decaying, the daughter isotope is often
radioactive and will decay so producing a granddaughter and so on. This is illustrated below:

Ra Rn + He
Radium Radon alpha particle
Parent daughter

This process is described as the transmutation of radium into radon with the emission of an alpha particle.

This then continues with Radon decaying as follows:

Rn Po + He
Radon Polonium alpha particle

The original parent radium atoms undergo a series of transmutations into new, radioactive daughter
elements ending finally with a daughter element that is stable. This is known as a radioactive decay series.
An illustration of the decay for U-235 is given on the following page.

66 Module 8 From the Universe to the Atom

Decay chain series
The following diagram shows the decay series for uranium-235. Note that the end product is lead-207 which
is stable.

Using the graph above write equations for the decay series of uranium 235. The first one has already been
done for you.

232 2301 ℎ + 42

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HSC Physics for NSW Study Guide Series
The properties of alpha, beta and gamma radiation
Your teacher may demonstrate the properties of alpha, beta and gamma radiation. As an alternative or
preparation for the demonstration search YouTube for Types of Nuclear Radiation by Fermilab and
answer the following questions. Open
Name four types of nuclear radiation.

What is alpha radiation and how is it produced?

When an alpha particle is emitted from 232 what are the two products?
What is beta radiation and how is it produced?

When a beta particle is emitted from 164 what are the two products?
What is gamma radiation and how is it produced?

When a gamma ray is emitted from 260 what are the products?
What is neutron radiation?
When a neutron is emitted from 2 2 what are the products?
How can you stop alpha radiation?
How can you stop beta radiation?
How can you stop (reduce) gamma radiation?
How can you stop neutron radiation?
Copy, on the next page, the diagram from the video showing the types of radiation and their penetrating
power.

68 Module 8 From the Universe to the Atom

Practical investigation penetrating power of radiation

Describe with the help of a labelled diagram an investigation to study the penetrating power of , ß and
radiation.

The differences in the penetrating power of radiation
As a radioactive particle passes through air it separates the air molecules into ions by removing an electron
and leaving behind a positive air molecule. We call the electron and the positive ion an ion pair.

path of radioactive particle

ion pairs

Alpha particles have a charge of plus two and travel relatively slowly. Because of this they are the most
strongly ionising and have the least penetrating power as they easily lose their energy. A gamma ray being
uncharged has a low ionising power and, because of this a high penetrating power.

Penetrating power
The following table shows the penetrating power of alpha ( ), beta (ß) and gamma ( ) rays that are produced
by naturally radioactive substances.

Radiation Range
Alpha Particle Stopped by paper
Beta Particle
Gamma Ray Stopped by aluminium
Reduced by lead

Other measurements show that the penetrating power is directly proportional to the inverse of the ionising
power.

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Magnetic fields and radiation
Beta particles are deflected like electrons in a magnetic field and are therefore negatively charged. Other
measurements such as the charge to mass ratio show that beta particles are electrons.
Alpha particles are deflected the opposite way in a magnetic field to beta particles, which means they are
positively charged. Other experiments show that an alpha particle is the nucleus of a helium atom.
Gamma rays are not deflected in a magnetic field therefore they have no charge. A gamma ray is part of
the electromagnetic spectrum as are light and X rays.
We now know that:

an particle is a helium nucleus
a ß particle is an electron
a ray is electromagnetic radiation.
Question sheet 8.9
1. There are three types of naturally occurring radiation.
(a) Describe an investigation to show the effect of magnetic fields on these radiations. You may need

to research further.

(b) Outline using a table the results you would expect for each of the three radiations.

2. Give a reason why:
(a) radioactive sources should not be touched with bare hands.

(b) radioactive sources emitting -rays should be stored in lead containers with thick sides.

3. All three types of radiation cause some ionisation of gases.
(a) Explain what is meant by the term ionisation of gases.

(b) Suggest a reason why -radiation produces very little ionisation.

(c) Suggest a reason why -radiation is very ionising.

70 Module 8 From the Universe to the Atom

4. Beryllium-9 was hit with alpha particles. One neutron per atom of beryllium was ejected. Write a
balanced equation for the reaction identifying the element that remained after the reaction.

5. Atoms of sulfur-32, on being hit and absorbing a neutron, emitted a proton. Write a balanced equation
for the reaction identifying the element that remained after the reaction.

6. A nuclide, , ejects an alpha particle, followed by two beta particles one after the other. Write the
three nuclear equations to represent the processes. Show that the final daughter nuclide is an isotope of
the first nuclide.

7. Using the graph below write equations for the decay series of thorium 232.

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Radioactive decay
It might appear that radioactive substances such as those in schools keep on emitting radiation forever but
they do lose activity over time. This loss in activity is related to the half-life of the radioactive source.
The half-life for a radioisotope is a measure of the tendency of the nucleus to decay and is based on
probability. The half-life is the time for one-half of the atomic nuclei of a radioactive sample to decay. The
radioactive isotope cobalt-60 has, for example, a half-life of 5.26 years. Hence after 5.26 years a sample
originally containing 32 grams of cobalt-60 would contain only 16 grams of cobalt-60 and would emit only
half as much radiation. After another interval of 5.26 years, the sample would contain only 8 grams of
cobalt-60 and so on.

Practical investigation alpha decay

Use the PhET alpha decay simulation. Click on Run Now and when the simulation opens click on the
Single Atom Tab. Watch the Polonium-211 nucleus until it decays. Click Reset Nucleus and watch it
again. Repeat this about ten times more. Openv
1. What difference do you find with the polonium-211 nucleus before and after it decays?

2. What particle comes from the nucleus? You can pause the simulation to have a closer look.

3. Why is there a change of mass number after a decay?

4. What does the polonium nucleus change into after it decays?

5. Write a balanced nuclear equation for this decay.

6. The decay time chart at the top of the screen shows the decayed nuclei of the previous trials you watched.
Is there a pattern for the decay times shown on the chart? Does the polonium decay with a predictable
pattern or does it appear random?

7. Click on Multiple Atoms and then click pause. Empty the Bucket o Polonium by clicking the Add
10 button until the bucket is empty. You should now have 100 atoms. Now click play and observe the
decay chart across the top as the atoms decay. What do you observe?

72 Module 8 From the Universe to the Atom

8. Click on Reset All Nuclei and observe again. Is the pattern of decayed nuclei across the top the same
as it was previously? Repeat this observation a few times to check your answer. Describe what you
observe.

9. Click on Reset All and choose custom nuclei. Move the green arrow until the half-life is over the 1.
Make sure the pause button is clicked and add 100 nuclei. Now click play but this time click the pause
button every five seconds. You may need another student with a stopwatch to help you take
measurements. Complete the table below, note the top left-hand side has a count of the atoms remaining
and those decayed. Repeat the investigation five times and calculate the average number of polonium-
211 nuclei remaining to the nearest whole number.

Number of polonium-211 nuclei remaining Average Time (seconds)
100, 100, 100, 100, 100 100 0
5
10
15
20
25
30

10.Plot a graph of this data with time on the horizontal axis and the average number of Polonium-211 nuclei
remaining on the vertical axis. Draw a smooth curve of best fit.

This shape of this graph is described as an exponential decay. We now look at the mathematics of this type
of relationship.

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The exponential decay relationship
The following section uses the exponential decay relationship, that is = − where:
= number of particles at time , 0 = number of particles present at t = 0, = decay constant,
1/2 = time for half the radioactive atoms to decay.
This relationship, = − produces a graph such as that shown below.
The exponential decay graph

2500
Number of particles
2000

1500

1000

500

0
01234567

time

Since the number of atoms present is proportional to the activity of the source, we can also write =
−
where: = the activity at time , 0 = the activity at t = 0, = decay constant, 1/2 = time for half the
radioactive activity to decay. Notice that the activity is directly proportional to the count rate. Activity is
measured in becquerels (Bq), where one becquerel is one nuclei decaying per second. The becquerel is
equivalent to an inverse second, s-1.

These equations say that the number of atoms left, or the activity, depends on:
how many you started with 0, or the activity you started with, 0
how quickly they decay, , that is the decay constant
how long you left them for, t.

We can derive another equation as follows:

Using the exponential decay relationship, = −
When the time t is 1/2 then half the atoms have decayed, hence Nt = ½N0. Substituting in we get:

1 0 = − or 1 = − or 2 = 1
2 2 −

We then take natural logs of both sides to get:

ln 2 = t1/2 or = 2 where t½ is the time for half the radioactivity to decay.

1/2

Note: ln 2 is the natural logarithm of 2. Check on your calculator that this gives the value of 0.693. This is

why we also see this relationship written as = 0.6 3 or 1/2 = 0.6 3.
1/2

74 Module 8 From the Universe to the Atom

Another relationship
The number of half-lives can be calculated using = where n is the number of half-lives that have

1/2

passed and t is the time in the same units as the half-life t½.

To find the radioactive nuclei remaining we use the relationship = (1)

1 2
2
If n = 1, one half-life has passed then =

If n = 2, two half-lives has passed then = 1
4

Example
Caesium-137 has a half-life of 30 years. If 100 g of caesium-137 disintegrates over a period of 120 years,
how many g of caesium-137 would be left?

In this case = = 120 = 4.

1/2 30

The mass remaining is given by = (1) . In this case Nt = 100 x (1)4 = 6.25 mg.
2
2

Want more information on these equations?
Search YouTube for Exponential decay formula proof Chemistry Khan Academy . Open
Note this is mathematical and not suitable or necessary for all students.

Research
An isotope that is frequently used in medicine is technetium-99m. This is manufactured at the OPAL nuclear
reactor in Sydney and has many uses. Find out its half-life and describe a few of its medical uses.

Another isotope frequently in the news is carbon-14. Find out its half-life and describe a few of its uses as
well as the nuclear equation for its decay.

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Practical investigation modelling half-life with M&M s

Safety precautions:
Do not eat the M&M s - they may have been sitting in the bag for some time and the laboratory is not a
hygienic place. Chocolate M&M s might be best, i.e. avoid peanuts. Students with allergies to any of the
ingredients in he M&M must be careful.

Instructions
Each group requires 20 M&M s and a ip lock bag. Consider all of the M&M s to be radioactive at the
start. Place the M&M s into the ip lock bag, shake the bag and then spill out the lollies onto the desktop.
Only the lollies with the m showing should still be considered radioactive, the others are assumed to have
decayed. Count and record the m lollies remaining and return to the bag. The lollies that are blank are
those that have decayed to a stable state and should be discarded .

Complete the table of results below.

Number of Other groups results Total number from all groups Throw number
radioactive M&M s
0
20 1
2
3
4
5

Now repeat the experiment recording your results and continue until all of the M&M s have decayed or
you have reached throw number 5.

Next, add the results of other groups into your table.

Questions
Did each group get the same results?

Why are the totals for all the groups better than individual group results?

Plot a graph of the number of radioactive M&M s on the vertical axis and throw number on the horizontal
axis.

76 Module 8 From the Universe to the Atom

Is the result a straight or a curved line? What does the line indicate about the nature of radioactive decay?

Suggest how you might improve this investigation. Also suggest an alternative simulation using something
other than M&M s.

Sample problem 8.6 Background radiation
A student takes measurements of background radiation using a Geiger-
Muller counter and records readings every 30 s. These are shown Time (s) Count
opposite.
00
(a) Calculate the value of the background radiation.
30 13

60 31

90 49

120 59

150 75

180 87

210 102

240 121

A radioactive source was then placed in front of the Counts per minute every hour
Geiger-Muller counter and readings of activity taken
for 1 minute at time intervals of 1 hour. The table Time (hour) Count Corrected
opposite gives these values. /minute count

(b) Draw a graph of activity against time. Thus, calculate rate/minute
the half-life of the source.
0 729 699
(c) Calculate the decay constant.
1 509 479
(d) The student continued looking at the measurements
after 7 hours and noticed a slight increase in the count 2 320 290
rate. Is this significant? Explain your answer.
3 201 171

4 135 105

5 95 65

6 70 40

7 50 20

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Activity (counts/minute)HSC Physics for NSW Study Guide Series
Solution
(a) The background can be calculated by taking the reading at the four-minute mark and dividing by 4 which

gives: 121/4 = 30 counts/minute. This background is then subtracted from the reading and entered into
the right-hand column of the table.
(b) A graph of activity against time is shown below.

900
800
700
600
500
400
300
200
100

0
012345678
time (minutes)

The half-life can be calculated by taking two points on the graph where the activity drops by one-half,
for example 600 and 300, and calculating the change in time.
The time at 600 counts per minute is 0.5 minutes.
The time at 300 counts per minute is 1.9 minutes.
The difference is 1.9 0.5 = 1.4 minutes. This is the half-life.
Note when you do this draw horizontal and vertical lines on your graph and show your working.
(c) The decay constant is given by = 2 = 0.6 = 8.21 x 10-3 s-1.

1/2 1.4 60

(d) The increase in count rate towards the end is not significant. As radioactivity is a random event we
would expect this type of behaviour.

78 Module 8 From the Universe to the Atom

Sample problem 8.7
Actinium-226 has a half-life of 29 hours. If 100 mg of actinium-226 disintegrates over a period of 87 hours,
how many mg of actinium-226 will remain?

Solution:

Two methods to solve this problem are:
The number of half-lives can be calculated using = where n is the number of half-lives that have

1/2

passed, t the time in the same units as the half-life t½.

In this case = = = 3.

1/2 2

The mass remaining is given by = (1) in this case 100 x (1)3 = 12.5 mg.

2 2

We can also use the relationship:

= − and = 2
1/2

First find the decay constant = 2 = 2 = 2.39 x 10-2 hr-1
1/2 2

Then substitute into = − this gives = 100 −2.3 10−2 = 12.5 mg

As you can see both methods give the same answer.

Question sheet 8.10
1. Caesium-137 has a half-life of 30 years. If 2.8 g of caesium-137 disintegrates over a period of 90 years,

how much caesium-137 would be left?

2. Sodium-25 was produced in a nuclear reactor but it took 4.0 minutes to get the sodium to the laboratory.
If 5.0 mg of sodium-25 was taken from the reactor, how many mg of sodium-25 arrived in the laboratory
4.0 minutes later if the half-life of sodium-25 is 60 seconds?

3. The half-life of an isotope is 2.0 years. How many years would it take for a 4.0 mg sample of the sample
to decay and have only 0.50 mg of it left?

4. Selenium-83 has a half-life of 25.0 minutes. How many minutes would it take for a 10.0 mg sample to
decay and have only 1.25 mg of it left?

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5. Polonium-211 decays to lead-207 with a decay constant of 1.343 s-1.

(a) Calculate the half-life of polonium-211.
(b) Calculate the percentage of polonium-211 remaining after 2.8 s.
6. Strontium-90 has a half life of 28.8 years.
(a) Calculate the decay constant of strontium-90.
(b) The Chernobyl disaster released strontium-90 into the environment. Strontium-90 is deposited in

bones and bone marrow. Outline why this is a problem.
7. A piece of wood from an ancient spear has a mass of 1 kg. An activity of 7.5 disintegrations per minute

is recorded from it (assume due to the decay of the isotope carbon 14). A similar modern replica made
from the same wood but with a mass of 2 kg has an activity of 30 disintegrations per minute. If the half-
life of carbon 14 is 5730 years calculate the age of the ancient spear.
8. Sketch and explain a graph that represents the decay of a radioactive isotope over time?

9. Neptunium-235 has a half-life of 400 days. Determine the radioactive decay constant.

80 Module 8 From the Universe to the Atom

Practical investigation nuclear fission simulation

Open the PhET nuclear fission simulation. Open
1. Select the tab Fission One Nucleus to answer the following:

(a) What can you do to make Uranium-235 unstable?

(b) How do you know it s unstable?

2. Describe what happens when a single nucleus of Uranium-235 is hit with a neutron. It may help to pause
the simulation.

3. Imagine that you have many U-235 nuclei and you fire a neutron at one of them. What do you think will
happen?

Now use the tab chain reaction .
Uncontrolled chain reaction
4. Increase the number of U-235 nuclei by increments of 25, each time firing a neutron. Discuss the

difference in the reaction rate as the number of U-235 increases.

5. Why is U-235 a good isotope for creating chain reactions?

6. Now reset all and repeat part 2 onwards with a uranium-238 nucleus and U-235 set to zero. What
difference does the uranium-238 make?

Controlled chain reaction
7. Explore the features of the Nuclear Reactor tab. What is the purpose of the control rods within a

nuclear reactor?

We can see here the connection to the inquiry question how can the energy of the atomic nucleus be
ha ne ed?

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Nuclear fission
We can release energy from nuclei if we split the nuclei of some heavy elements (uranium-235 or
plutonium-239) into two approximately equal parts. We can split a uranium-235 nucleus if a neutron hits it
and is captured by the uranium nucleus. For example, with uranium-235, common mass numbers of the
fission fragments are 93 and 140. This nuclear fission is illustrated below.

neutron caesium
uranium neutron
neutron
neutron

rubidium

The reaction is: 10 + 232 → 140 + 33 + 3 01 + energy

Chain reaction
The three neutrons released can go on to hit further U-235 nuclei keeping the reaction going. This is called
a chain reaction and is illustrated below.

neutron 3 neutrons
uranium 3 neutrons

3 neutrons

Controlled and uncontrolled nuclear chain reactions
A chain reaction refers to a process in which neutrons are released in fission to produce an additional fission
in at least one further nucleus. This nucleus in turn produces neutrons, and the process repeats. The process
may be controlled, as in the case of a nuclear reactor, or uncontrolled as in the case of nuclear weapons.

Controlled nuclear chain reaction
To maintain a sustained controlled reaction, for every 2 or 3 neutrons released, only one must be allowed
to hit another uranium nucleus. If this ratio is less than one, then the reaction will stop.

absorbed neutron neutron absorber

neutron
uranium

absorbed neutron

Control of the chain reaction is managed by controlling the number of neutron collisions that produce
fission. This can be done by:

absorbing neutrons with a neutron-absorbing material such as cadmium
letting neutrons escape or
separating the fissionable material.

Australia s only nuclear reactor is the Open Pool Australian Lightwater (OPAL) reactor situated at Lucas
Heights, 40 km from the Sydney CBD. It does not produce electricity instead being used to produce isotopes
for nuclear medicine amongst many other uses.

82 Module 8 From the Universe to the Atom

Uncontrolled nuclear chain reaction
To produce an uncontrolled nuclear reaction, for every 2 or 3 neutrons released, more than one must be
allowed to hit another uranium nucleus. This means that it produces an uncontrolled nuclear reaction; an
explosion.

neutron
uranium

Modelling an uncontrolled nuclear chain reaction
Students are asked to model a chain reaction. Watch the following YouTube video, search for Mousetrap
Fission Harvard Natural Sciences Lecture Demonstrations and discuss the success of this model.

Find another method of modelling a chain reaction and outline below.

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The release of energy in radioactive decay
Nuclei which are radioactive become more stable by emitting radioactivity in the form of alpha, beta or
gamma decay. We now look at some examples of this:

Alpha decay
An example is the decay of radon-222, the reaction is: 2262 → 214 + 42
Table of masses:

Particle Atomic mass (u)
Rn-222 222.017570
Po-218 218.008973
alpha particle 4.001506

The mass defect is given by (mass of reactants - mass of products)

That is (222.017570) (218.008973 + 4.001506) = 7.09 x 10-3 u and as 1 u has 931.5 MeV of energy then
the energy released is given by 7.09 x 10-3 x 931.5 MeV = 6.6 MeV.

When this reaction is studied it is discovered that the kinetic energy of the Alpha particle is about 5.5 MeV.
Not all the kinetic energy goes in the alpha particle, some will also be given to the polonium as it must have
velocity to satisfy the conservation of momentum.

Beta decay
An example is the decay of carbon-14, the reaction is: 164 → 14 + −01 +

The mass defect given by the mass of reactants - mass of products is converted into mainly kinetic energy
of the beta particle. The antineutrino is necessary to satisfy the conservation of energy and momentum. We
discuss the antineutrino later when looking at the standard model of matter.

Gamma decay
Gamma decay produces no change in the number of protons and neutrons however the nucleus does release
energy in the form of a gamma ray and so is more stable.

Gamma decay also occurs with the annihilation of a positron with an electron, in this case two gamma rays
are produced. +01 + −01 → 2 . A calculation for this follows:

Table of masses:

Particle Atomic mass (u)
positron 9.109 x 10-31 kg
beta particle 9.109 x 10-31 kg

The energy released is given by E = mc2 = 2 x 9.109 x 10-31 x (3 x 108)2 J = 1.64 x 10-13 J = 1.02 MeV.
This energy is shared between the gamma rays giving them each an energy of about 511 keV.

84 Module 8 From the Universe to the Atom

The release of energy in the process of nuclear fission
The energy released during the fission process can be explained from the conservation of mass-energy.
During the fission process there is a decrease in the mass of the system. Therefore, there must be energy
released equal to the energy equivalence of the mass lost in the process.

The masses in the table below are for the following fission reaction.
10 + 232 → 140 + 33 + 3 10 + energy

Isotope Atomic mass (u)
U-235 235.043924
Cs-140 139.909100
Rb-93 92.916990
neutron 1.008665

The mass defect is given by (mass of reactants - mass of products)
That is (235.043924 + 1.008665) (139.909100 + 92.916990 + 3 x 1.008665) = 0.200509 u
The energy released is given by 0.200509 x 931.5 MeV = 186.8 MeV.

The release of energy in the process of nuclear fusion

A typical fusion reaction is when two deuterium nuclei join together to form tritium, that is hydrogen-3,

and a proton. The reaction is as follows: 2 + 2 → 13 + 11
1 1

Using the masses in the table below we can calculate the energy released.

nuclei Atomic mass (u)
deuterium 2.014102
tritium 3.016049
proton 1.007276

The mass defect is given by (mass of reactants - mass of products)
That is (2 x 2.014102) (3.016049 + 1.007276) = 4.879 x 10-3 u
The energy released is given by 4.879 x 10-3 x 931.5 MeV = 4.5 MeV.

Note: the values given for atomic mass depend upon the source used and whether it was a nuclide or an
atom, that is with electrons. You will be given the appropriate values in any questions.

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Binding energy
A nuclear reaction releases energy as the products are more stable than the reactants, that is the mass of the
products is less than the reactants causing energy release from Einstein s relationship, E = mc2. As well as
using mass defect to study the energy released in fission binding energy can also be used.

As an introduction to binding energy watch the YouTube video search for Binding Energy per Nucleon
Daniel M Open. Watch the video through without taking notes then watch again answering the following.

When a proton and neutron join together what do they emit?

Why do bigger nuclei have a smaller than expected mass? ....
.

The bigger the magnet the bigger the ..

The bigger the force the bigger the ..

The smaller the nucleus the .. the nuclear force.

The bigger the nucleus the .. the nuclear force.

As a nucleon joins a nucleus what is emitted? ..

Why does a very large nucleus not have the greatest binding energy?

Why does a medium sized nucleus lose more binding energy than a very large nucleus?

Sketch the binding energy per nucleon graph shown at 6.00 minutes into the video.

86 Module 8 From the Universe to the Atom

Another video to watch on binding energy is Nuclear Binding Energy tutorial (Post 16 physics) Open.
This covers the same material but with a different emphasis. Watch the video through without taking notes
then watch again answering the following.
What holds the particles in the helium-4 nucleus together?
What is the definition of binding energy?
Draw the weighing scales with the He-4 nucleus as shown at the 1.5 minute mark in the video.

Show the steps/values to get the binding energy/nucleon value for helium-4.

What is special about Fe-56 on the binding energy per nucleon graph?

Module 8 From the Universe to the Atom 87

Binding energy per nucleon (MeV)HSC Physics for NSW Study Guide Series
Energy releasedBinding energies
by fusionThe graph below shows the binding energy per nucleon against nucleon number. Elements with a high
binding energy per nucleon are very difficult to break up. Iron 56 with a binding energy/nucleon of about
8.8 MeV is very stable with the highest binding energy per nucleon of any element and this explains why
there is so much of it in the universe.

9

8

7

6

Energy released by fission

5

4

3

2

1

0 20 40 60 80 100 120 140 160 180 200 220 240

mass number (A)

The part of the curve to the left shows that two light elements can produce energy by joining together, that
is fusion, while the part of the curve to the right shows that a heavy element can produce energy by splitting,
that is fission.

For example, when we have the alpha decay of radon-220, the reactants polonium-216 and an alpha particle
have more total binding energy and consequently move up the binding energy per nucleon curve, causing
energy to be released.

The helium nucleus has a high binding energy per nucleon and is more stable than some of the other nuclei
close to it in the periodic table. This is shown in the table as mass number four with binding energy per
nucleon of 7.1 MeV.

Sample problem 8.8
The mass of the arsenic-75 nucleus is 74.9216 u. Take the mass of the proton as 1.007276 u and the neutron
as 1.008664. Calculate the total binding energy and binding energy per nucleon for this nucleus.

Solution:
Looking at the periodic table we see that the atomic number of arsenic is 33. This means that arsenic-75
has 33 protons and (75-33) = 42 neutrons. This gives:
protons 33 x 1.007276 u = 33.24011
neutrons 42 x 1.008664 u = 42.36389

Total = 75.6040

The mass defect is 75.6040 - 74.9216 = 0.6824 u
1 u = 931.5 MeV hence the total binding energy is 0.6824 x 931.5 = 635.7 MeV.
The binding energy per nucleon is 635.7/75 = 8.5 MeV/nucleon.

If you look at the binding energy per nucleon graph you will see that this value is in agreement.

88 Module 8 From the Universe to the Atom

Question sheet 8.11
1. Explain why energy is released when the nuclei of a heavy element fissions.

2. Explain why energy is released when two light nuclei fuse together.

3. The following information is necessary to answer this question.

Isotope Atomic mass (u) Name Number of protons
La-139 138.9063
Mo-95 94.9058
U-235 235.0439

(a) Complete the table giving the elements La and Mo their name and number of protons.

(b) U-235 and a neutron react to produce Mo-95, La-139, 2 neutrons and a quantity of another
subatomic particle. Identify the subatomic particle by writing a balanced nuclear equation.

.

(c) Calculate the total energy released in MeV and Joules for this reaction.

4. A neutron collides with a U-235 nucleus. These reactants split into Ba-141, Kr-92 and three neutrons.
(a) Write the nuclear equation for this reaction.

(b) Estimate the energy released in this fission reaction using the binding energy per nucleon values
given in the graph on the last page.

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Inquiry question: How can the energy of the atomic nucleus be harnessed?

Now that you have covered this section you should be better able to answer the inquiry question with your new
understanding of classical physics and why it cannot explain the properties of the atom.

Summary
Write a summary of properties of the nucleus, review your summary with others in the class, update as
necessary.

.
.
.
.
.
.
.
.
.
.

90 Module 8 From the Universe to the Atom

Deep inside the atom

Students:
analyse the evidence that suggests:
- that protons and neutrons are not fundamental particles
- the existence of subatomic particles other than protons, neutrons and electrons

investigate the Standard Model of matter, including:
- quarks, and the quark composition hadrons
- leptons
- fundamental forces (ACSPH141, ACSPH142)

investigate the operation and role of particle accelerators in obtaining evidence that tests and/or validates
aspects of theories, including the Standard Model of matter (ACSPH120, ACSPH121, ACSPH122,
ACSPH146)

NSW Physics Stage 6 syllabus © NSW Education Standards Authority for and on behalf of the Crown in right of the State of New South
Wales, 2017.

Inquiry question: How is it known that human understanding of matter is still incomplete?

With all inquiry questions consult with class members, come to a consensus and then summarise your answer in the
space provided.

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Introduction
Search YouTube for Standard Model - Fermilab Open. Watch the video through without taking notes
then watch again answering the following.

What device during the 1940-60s helped physicists to discover many subatomic particles? .
Complete the following table with quark name and any other information about the particle.

Quark Information

Name two well-known particles which are made from up and down quarks. .
Name another sub-atomic class of particles. .
Name a well-known particle which belongs to this sub-atomic class. .
Complete the following table of the four fundamental forces and what we know about them.

Force Information about force The range of the force Relative strength of the force

Unsolved mysteries
This video was produced just before the discovery of the Higgs boson details of which follows.

The Higgs boson
The standard model of particle physics is the model used by physicists to understand the elementary
constituents of all matter and the forces that hold matter together, or cause it to fall apart. The standard
model is not yet a theory of everything. It does not account for the force of gravity.
The Higgs boson is important in the standard model because it suggests the existence of a Higgs field, a
force field which is found throughout the universe. Without the Higgs field, particles such as quarks and
electrons would be massless and travel at the speed of light. The Higgs field causes elementary particles to
interact with each other and to slow down, as though swimming in treacle. The Higgs boson was proposed
by British physicist Peter Higgs and others in 1964 but was not confirmed until July 2012. From this we
can see that the fundamental particle model is forever being updated.

92 Module 8 From the Universe to the Atom

Other resources
Nuclear physics chart found at www.cpepphysics.org/images/2014-fund-chart.pdf
Your teacher may be able to print this out for your classroom wall in colour and in A3 format.

Website and phone app
Investigate the offerings at www.particleadventure.org/accel.html
This is also available as an app for your phone. Search for the Particle Adventure Berkeley Lab.

Evidence that protons and neutrons are not fundamental
Until the start of the 20th century atoms were thought to be the fundamental building blocks of matter and
were not divisible. Experiments by Rutherford and others however showed that the atom was in fact
divisible and mostly empty space with electrons surrounding a nucleus made up of protons and neutrons.
With the invention of particle accelerators that accelerated protons and electrons to high energies it was
discovered that there existed a very large number of new particles, that is protons and neutrons were not
fundamental. By the early 1960s, as accelerators reached higher energies, a hundred or more types of
particles were discovered.

Background
To investigate deep inside the atom high energy particles need to be smashed into the nucleus. High speeds
are required to overcome coulombic repulsion. Up until 1932 only alpha particles could be used to study
nuclear reactions and these could only be obtained in beams of low intensity and kinetic energy. The
development of accelerators allowed particles to be given much higher energies than those from naturally
occurring alpha particles.
Particle accelerators are used to accelerate protons and other charged particles. The speeds of the
accelerated particles are so high that their mass increases due to relativistic effects. The particles are then
allowed to collide with other particles to form new products which are very unstable. These new products
are then detected and analysed to determine the structure of matter in terms of fundamental particles and
their interactions. These fundamental particles and their interactions form the basis of the Standard Model.

Research task particle accelerators
Students have to investigate the operation and role of particle accelerators in obtaining evidence that
tests and/or validates aspects of theories, including the Standard Model of matter.

Some possibilities to investigate include:

Fermilab - America's particle physics and accelerator laboratory.

The Large Hadron Collider (LHC) - the world's largest and most powerful particle collider and the
largest machine in the world. It lies in a tunnel 27 kilometres in circumference beneath the France
Switzerland border.

As an introduction to particle accelerators search YouTube for How does an atom-smashing particle
accelerator work? by Don Lincoln. Open

Summarise your task making sure you answer the requirements given above in bold/italics and do not just
rewrite the information given in web pages. Get another student to read your work and provide constructive
criticism. Once this has been done and your work improved/modified write out on the following page.

Module 8 From the Universe to the Atom 93

HSC Physics for NSW Study Guide Series

94 Module 8 From the Universe to the Atom

The standard model
The standard model is the latest model of matter developed by using powerful particle accelerators such as
the Large Hadron Collider (LHC). The standard model divides fundamental particles into two main groups:

Fermions the particles which make up matter
Bosons the particles which mediate, that is bring about force.

Fermions

Fermions are particles that make up matter such as quarks and leptons as well as composite particles like

protons and neutrons.

Quarks

There are six quarks: Up, Down, Charm, Strange, Top and Bottom.

Quarks have a fractional charge of either + 2 or - 1 compared to the negative one charge of the electron
3 3
and the plus one charge of the proton. The table below shows the different types of quarks and their charge.

Quark name Charge relative to size of proton charge
up (u)
down (d) + 2
strange (s)
charm (c) 3
bottom (b)
top (t) - 1
3

- 1
3

+ 2

3

- 1
3

+ 2
3

Quarks also have antimatter equivalents, antiquarks, represented by a line over the letter, for example ̅ is
the antimatter up quark. The charge on an antiquark is equal but opposite to that of the normal quark.

Hadrons
Fractional charges like those of quarks have not been observed because quarks are never found separately,
but only in combined particles called hadrons. A hadron is a particle made of a combination of two or
more quarks or antiquarks. There are two classes of hadrons, the baryon and the meson. The following
diagram shows some hadrons.

Hadrons

Baryons Mesons

u d
d du su

u neutron (udd) negative kaon positive kaon

proton (uud)

Module 8 From the Universe to the Atom 95