How many QCAA Physics Alternative Sequence questions cover Ionising radiation and nuclear reactions?

AusGrader has 53 QCAA Physics Alternative Sequence questions on Ionising radiation and nuclear reactions, all with instant AI grading and detailed marking feedback.

QCAA Physics Alternative Sequence — Ionising radiation and nuclear reactions Questions & Answers

Q21

2025

NESA

3 marks

Q21

3 marks

A scientist has two unlabelled sources of radiation. One source emits alpha particles and the other emits beta particles.

Outline TWO methods that could be used to determine which source is the alpha emitter, and which source is the beta emitter.

Reveal Answer

Pass the radiation through a thin material to a detector. Observe any changes in counts. The beta source count will be reduced by an amount less than the alpha source count.

Pass each type of radiation through a magnetic field and observe any differences in deflection and observe deflection in opposite directions.

Marking Criteria

Descriptor	Marks
Outlines TWO methods and subsequent observations that could be used to identify each source as either alpha or beta emitters	3
Outlines a method that could be used to distinguish the sources OR Identifies TWO methods that could be used to identify each source	2
Provides some relevant information	1
None of the above	0

Q13

2023

NESA

1 mark

Q13

1 mark

Nucleus X has a greater binding energy than nucleus Y.

What can be deduced about X and Y?

A

X is more stable than Y.

B

Y is more stable than X.

C

X has a greater mass defect than Y.

D

Y has a greater mass defect than X.

Reveal Answer

A

X is more stable than Y.

Stability is determined by binding energy per nucleon, not total binding energy, so we cannot determine which nucleus is more stable without knowing their nucleon numbers.

B

Y is more stable than X.

Stability depends on binding energy per nucleon, which cannot be determined from total binding energy alone.

C

X has a greater mass defect than Y.

Correct Answer

Binding energy is directly proportional to mass defect according to the mass-energy equivalence equation $E = \Delta m c^2$ . Therefore, a greater binding energy means a greater mass defect.

D

Y has a greater mass defect than X.

Since nucleus X has a greater binding energy, it must have a greater mass defect than Y, not the other way around.

Q31

2025

NESA

5 marks

Q31

5 marks

Experiments have been carried out by scientists to investigate cathode rays.

Assess the contribution of the results of these experiments in developing an understanding of the existence and properties of electrons.

Reveal Answer

Experimental results and their interpretation have played an essential role in developing and understanding of both the existence and properties of electrons.

It was demonstrated that cathode rays could be deflected by an electric field in a manner consistent with negatively charged particles. This experiment also allowed the deduction that the cathode rays were not electromagnetic waves because the latter would not be deflected by an electric field.

The combined effect of changing the magnitudes of electric and magnetic fields through which the electrons passed was measured and from this the charge-to-mass ratio of the electron was calculated.

Marking Criteria

Descriptor	Marks
Provides a comprehensive assessment of the experimental results Relates the results to both existence and properties of electrons	5
Provides a sound assessment of the experimental results Relates the results to the properties and/or existence of electrons	4
Outlines an experiment and/or results Relates experimental results to the existence and/or properties of electrons	3
Outlines a relevant experiment or result OR Relates an experimental result to the existence or property of electrons	2
Provides some relevant information	1
None of the above	0

Q1

2021

QCAA

Paper 2

3 marks

Q1

3 marks

Explain how an excess of positive charge in the nucleus causes beta positive radiation.

Reveal Answer

An excess of protons causes instability in the nucleus. To gain stability, one proton will change to a neutron. When this occurs, a beta positive particle is emitted from the nucleus.

Marking Criteria

Descriptor	Marks
identifies instability from an excess of positive charge as the cause for decay	1
identifies the decay of one proton into a neutron	1
identifies that a beta positive particle is emitted from the nucleus	1

Q3

2021

QCAA

Paper 1

1 mark

Q3

1 mark

Identify the correct formula for the mass–energy equivalence relationship.

A

$E = mc^2$

B

$E = mgh$

C

$E = \frac{1}{2}mc^2$

D

$E = \frac{1}{2}mv^2$

Reveal Answer

A

$E = mc^2$

Correct Answer

Correct. This is Einstein's famous mass-energy equivalence formula, stating that energy ( $E$ ) equals mass ( $m$ ) times the speed of light squared ( $c^2$ ).

B

$E = mgh$

Incorrect. This is the formula for gravitational potential energy, where $g$ is the acceleration due to gravity and $h$ is height.

C

$E = \frac{1}{2}mc^2$

Incorrect. This incorrectly combines the structure of the kinetic energy formula with the speed of light ( $c$ ), which does not represent any standard physical relationship.

D

$E = \frac{1}{2}mv^2$

Incorrect. This is the classical formula for kinetic energy, representing the energy of an object in motion with mass $m$ and velocity $v$ .

Q16

2025

NESA

1 mark

Q16

1 mark

A neutron is absorbed by a nucleus, $X$ .

The resulting nucleus undergoes alpha decay, producing lithium-7.

What is nucleus $X$ ?

A

Boron-10

B

Boron-11

C

Lithium-6

D

Lithium-10

Reveal Answer

A

Boron-10

Correct Answer

Working backwards, Lithium-7 ( ${}^{7}_{3}\text{Li}$ ) plus an alpha particle ( ${}^{4}_{2}\alpha$ ) gives an intermediate nucleus of Boron-11 ( ${}^{11}_{5}\text{B}$ ). Since this was formed by absorbing a neutron ( ${}^{1}_{0}\text{n}$ ), the original nucleus $X$ must be Boron-10 ( ${}^{10}_{5}\text{B}$ ).

B

Boron-11

If Boron-11 absorbed a neutron, it would form Boron-12. An alpha decay from Boron-12 would produce Lithium-8, not Lithium-7.

C

Lithium-6

If Lithium-6 absorbed a neutron, it would form Lithium-7 directly, but the question states that Lithium-7 is produced via alpha decay after the neutron absorption.

D

Lithium-10

If Lithium-10 absorbed a neutron, it would form Lithium-11. An alpha decay from Lithium-11 would produce Hydrogen-7, not Lithium-7.

Q13

2020

VCAA

1 mark

Q13

1 mark

Matter is converted to energy by nuclear fusion in stars.

If the star Alpha Centauri converts mass to energy at the rate of 6.6 × 10^9 kg s−1, then the power generated is closest to

A

2.0 × 10^18 W

B

2.0 × 10^18 J

C

6.0 × 10^26 W

D

6.0 × 10^26 J

Reveal Answer

A

2.0 × 10^18 W

Incorrect. This value incorrectly calculates power by multiplying the mass rate by the speed of light ( $c$ ) instead of $c^2$ .

B

2.0 × 10^18 J

Incorrect. This option uses the incorrect formula ( $mc$ instead of $mc^2$ ) and the wrong unit for power, which should be Watts, not Joules.

C

6.0 × 10^26 W

Correct Answer

Correct. Power is the rate of energy generation, calculated using $P = \frac{dm}{dt}c^2$ . Multiplying $6.6 \times 10^9 \text{ kg/s}$ by $(3.0 \times 10^8 \text{ m/s})^2$ gives approximately $6.0 \times 10^{26} \text{ W}$ .

D

6.0 × 10^26 J

Incorrect. Although the numerical calculation is correct, the unit for power is Watts (Joules per second), not Joules.

Q3

2023

QCAA

Paper 1

1 mark

Q3

1 mark

The radioactive decay of 16 g of sodium-24 was recorded over time.

Predict how many half-lives it would take before only 1 g of sodium-24 remained.

A

4

B

5

C

6

D

8

Reveal Answer

A

4

Correct Answer

This is correct because each half-life reduces the mass by half. Starting with $16\text{ g}$ , it takes 4 half-lives to reach $1\text{ g}$ ( $16 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1$ ).

B

5

This is incorrect. After 5 half-lives, the remaining mass would be halved one more time, leaving only $0.5\text{ g}$ of sodium-24.

C

6

This is incorrect. After 6 half-lives, the remaining mass would be $16 \times (1/2)^6 = 0.25\text{ g}$ , not $1\text{ g}$ .

D

8

This is incorrect. After 8 half-lives, the remaining mass would be $16 \times (1/2)^8 = 0.0625\text{ g}$ .

Q19

2025

NESA

1 mark

Q19

1 mark

A system consists of a sealed glass jar containing some oxygen and a small strip of magnesium.

The magnesium reacts with the oxygen to produce magnesium oxide as a product. Energy is released from the system in this reaction.

The mass of the system will

A

increase because oxygen is added to the magnesium.

B

decrease because energy is removed from the system.

C

increase because energy is added to the system by the reaction.

D

decrease because magnesium and oxygen are lost in the reaction.

Reveal Answer

A

increase because oxygen is added to the magnesium.

Incorrect. The oxygen is already part of the sealed system, so combining it with magnesium does not add new mass to the system as a whole.

B

decrease because energy is removed from the system.

Correct Answer

Correct. According to mass-energy equivalence ( $E = mc^2$ ), the release of energy from the system results in a proportionally tiny, but real, decrease in the system's total mass.

C

increase because energy is added to the system by the reaction.

Incorrect. The problem explicitly states that energy is released from the system, not added to it, which would decrease rather than increase the mass.

D

decrease because magnesium and oxygen are lost in the reaction.

Incorrect. The magnesium and oxygen atoms are not lost; they are simply rearranged into magnesium oxide within the sealed jar, conserving the total number of atoms.

Q24

2023

QCAA

Paper 1

2 marks

Q24

2 marks

Nitrogen-14 is a product of the nuclear decay reaction shown.

^{A}_{Z}X \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}e + \nu

Determine the starting isotope.

Reveal Answer

A nucleus undergoing beta negative decay experiences neutron decay into a proton and electron. Therefore

Z_{Nitrogen} - 1 = 7 - 1 = 6

The starting isotope is $^{14}_{6}C$

Marking Criteria

Descriptor	Marks
correctly determines atomic number and mass number	1
identifies element for calculated atomic number	1

Q3

2023

QCAA

Paper 2

3 marks

Q3

3 marks

Compare artificial transmutation and natural radioactive decay.

Reveal Answer

Both artificial transmutation and natural radioactive decay release energy as the nucleus of the atoms involved form new products as a result of a nuclear reaction.

While natural radioactive decay occurs spontaneously (without human intervention), artificial transmutation is a human-made process where specific particles are used to bombard specific nuclei to produce specific products.

The significance is that these neutrons can then go on to bombard other nuclei and cause them to split in the same way repeatedly (causing a chain reaction), which would cause artificial transmutations to release more energy than natural radioactive decay.

Marking Criteria

Descriptor	Marks
identifies a similarity	1
identifies a difference	1
identifies a significance	1

Q21

2023

QCAA

Paper 1

2 marks

Q21

2 marks

Describe what happens to a nucleus, in terms of particles and mass, when it undergoes beta positive decay.

Reveal Answer

The number of nucleons is not changed during beta positive decay because a proton becomes a neutron. The positron is of negligible mass. Therefore, there is no change in the atomic mass of the atom.

Marking Criteria

Descriptor	Marks
describes that nucleon number does not change	1
identifies no significant change in atomic mass of nucleus	1

Q4

2023

SCSA

4 marks

Q4

4 marks

Calculate the wavelength of a photon with an energy of 1.81 keV.

Reveal Answer

First convert the energy to joules:

\begin{align*} E &= 1.81\ \text{keV} \\ &= 1810\ \text{eV} \\ &= 1810(1.60\times10^{-19}) \\ &= 2.896\times10^{-16}\ \text{J} \end{align*}

Using $E = hf$ and $f = \frac{c}{\lambda}$ ,

\begin{align*} E &= \frac{hc}{\lambda} \end{align*}

Rearranging for $\lambda$ :

\begin{align*} \lambda &= \frac{hc}{E} \\ &= \frac{(6.63\times10^{-34})(3.00\times10^8)}{2.896\times10^{-16}} \\ &= 6.87\times10^{-10}\ \text{m} \end{align*}

Marking Criteria

Descriptor	Marks
Converts to joules	1
Substitutes $\frac{c}{\lambda}$ for $f$	1
Rearranges for $\lambda$	1
Calculates answer	1

Q4

2023

NESA

1 mark

Q4

1 mark

Caesium-137 has a half-life of 30 years.
What mass of caesium-137 will remain after 90 years, if the initial mass was 120 g?

A

4 g

B

15 g

C

40 g

D

60 g

Reveal Answer

A

4 g

This incorrect answer comes from dividing the initial mass by the half-life ( $120 / 30 = 4$ ), rather than calculating the exponential decay over 3 half-lives.

B

15 g

Correct Answer

Since 90 years is exactly 3 half-lives ( $90 / 30 = 3$ ), the initial mass is halved 3 times. The remaining mass is $120 \text{ g} / 2^3 = 15 \text{ g}$ .

C

40 g

This incorrect answer comes from dividing the initial mass by the number of half-lives ( $120 / 3 = 40$ ), instead of dividing by $2^3$ .

D

60 g

This is the mass that would remain after only 1 half-life (30 years), not the 3 half-lives (90 years) specified in the question.

Q2

2021

SCSA

3 marks

Q2

3 marks

Using information from the Formulae and Data Booklet, calculate the mass of a bottom quark in kg.

Reveal Answer

Using the value from the data sheet of $4.18 \text{ GeV}/c^2$ , convert eV to Joules to get $4.18 \times 10^9 \times 1.6 \times 10^{-19} = 6.69 \times 10^{-10} \text{ J}$ .

Then the mass is $6.69 \times 10^{-10} / 9 \times 10^{16} = 7.43 \times 10^{-27} \text{ kg}$ .

Marking Criteria

Descriptor	Marks
Uses correct value from data sheet ( $4.18 \text{ GeV}/c^2$ )	1
Converts eV to Joules ( $4.18 \times 10^9 \times 1.6 \times 10^{-19} = 6.69 \times 10^{-10} \text{ J}$ )	1
Correctly calculates mass ( $6.69 \times 10^{-10} / 9 \times 10^{16} = 7.43 \times 10^{-27} \text{ kg}$ )	1

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