VCAA Physics How has understanding about the physical world changed?

15 sample questions with marking guides and sample answers

Q6
2022
QCAA
Paper 1
1 mark
Q6
1 mark

After coherent light has been passed through a double slit, the observation of an interference pattern on a screen is explained by the

A

wave nature of light.

B

equal width of the slits.

C

discrete packets of photons.

D

distance from the slits to the screen.

Reveal Answer
A

wave nature of light.

Correct Answer

Interference is a fundamental property of waves where overlapping wavefronts add constructively or destructively; this experiment is the classic evidence for the wave nature of light.

B

equal width of the slits.

While slit width affects the diffraction envelope and contrast, the interference pattern itself is caused by wave superposition, which can occur even if the slits are not perfectly equal in width.

C

discrete packets of photons.

Discrete packets (photons) refer to the particle nature of light; if light behaved strictly as classical particles, it would form two distinct bands rather than an interference pattern.

D

distance from the slits to the screen.

The distance to the screen affects the spacing of the fringes (scale), but it is not the fundamental cause of the interference phenomenon itself.

Q5
2023
QCAA
Paper 2
4 marks
Q5
4 marks

Describe what happens when light is shone onto a metallic surface in the context of the photoelectric effect.

Reveal Answer

Light with energy equivalent to hfhf has the ability to produce photoelectrons from a metallic surface.

When the frequency of light is below the threshold frequency for the metallic surface, the light will be reflected with no transfer of energy.

When the frequency of light is above the threshold frequency for the metallic surface, the energy of the photons will be absorbed and photoelectrons with kinetic energy proportional to the excess energy will be released.

The intensity of incident light is proportional to the number of photoelectrons for frequencies greater than the threshold frequency.

Marking Criteria
DescriptorMarks

Identifies incident light has energy equivalent to hfhf

1

Describes transfer of energy when frequency of light is below the threshold frequency

1

Describes transfer of energy when frequency of light is above the threshold frequency

1

Identifies relationship between intensity of incident light and resultant photoelectrons

1
Q10
2023
SCSA
4 marks
Q10
4 marks

Estimate the de Broglie wavelength for a standard men's basketball travelling at 10.0 m s1^{-1}.

Reveal Answer

Using de Broglie's equation,

λ=hmv\begin{align*} \lambda &= \frac{h}{mv} \end{align*}

Taking the mass of a standard men's basketball as approximately

m=0.62 kg\begin{align*} m &= 0.62\ \text{kg} \end{align*}

then

λ=6.63×1034(0.62)(10.0)=1.07×1034 m\begin{align*} \lambda &= \frac{6.63\times10^{-34}}{(0.62)(10.0)} \\ &= 1.07\times10^{-34}\ \text{m} \end{align*}

So the estimated de Broglie wavelength is approximately

λ1.1×1034 m\begin{align*} \lambda &\approx 1.1\times10^{-34}\ \text{m} \end{align*}
Marking Criteria
DescriptorMarks

Estimates mass of basketball

1

Substitutes mvmv for pp in equation (using 0.60 kg)

1

Calculates answer

1

2 significant figures

1
Q7
2023
QCAA
Paper 2
5 marks
Q7
5 marks

Discuss the nature of light by describing evidence from two key experiments.

Reveal Answer

Young's double slit experiment and black-body radiation both provide evidence for the behaviour of light.

In Young's double slit experiment, the interference patterns formed as light passed between the two slits demonstrates the wave nature of light.

In contrast, black-body radiation demonstrates the quantised nature of light as electrons can only absorb or emit energy in discrete amounts.

Therefore, light has some wave properties and some particle properties.

Marking Criteria
DescriptorMarks

Identifies evidence for the nature of light comes from Young's double slit experiment

1

Identifies evidence for the nature of light comes from black-body radiation

1

Describes evidence for wave nature of light

1

Describes evidence for photons

1

Concludes light has the properties of both waves and particles

1
Q10
2023
VCAA
6 marks
Q10

A proton in an accelerator beamline of proper length 4.80 km4.80 \text{ km} has a Lorentz factor, γ\gamma, of 2.002.00.

Q10a
3 marks

Calculate the speed of the proton relative to the beamline in terms of cc, the speed of light in a vacuum. Give your answer to three significant figures.

Reveal Answer

γ=11v2c2\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

... and after much transposition ...

v=c11γ2v = c \sqrt{1 - \frac{1}{\gamma^2}}

v=c1122v = c \sqrt{1 - \frac{1}{2^2}}

v=c×0.866=0.866cv = c \times 0.866 = 0.866c

Marking Criteria
DescriptorMarks

Identifies the correct formula or correctly transposes the Lorentz factor formula to solve for velocity, e.g., v=c11γ2v = c \sqrt{1 - \frac{1}{\gamma^2}}

1

Correctly substitutes the given values into the formula, e.g., v=c1122v = c \sqrt{1 - \frac{1}{2^2}}

1

Calculates the correct answer to three significant figures, 0.866c0.866c

1
Q10b
1 mark

Calculate the length of the beamline in the reference frame of the proton.

Reveal Answer

L=L0γL = \frac{L_0}{\gamma}

L=4.82.0L = \frac{4.8}{2.0}

L=2.4 kmL = 2.4 \text{ km}

Marking Criteria
DescriptorMarks

Calculates the correct length of the beamline in the reference frame of the proton, 2.4 km2.4 \text{ km}

1
Q10c
2 marks

Calculate the kinetic energy of the proton in joules. Show your working.

Mass of proton =1.67×1027 kg= 1.67 \times 10^{-27} \text{ kg}.

Reveal Answer

Ek=(γ1)mc2E_k = (\gamma - 1)mc^2

Ek=(21)×(1.6×1027)×(3.0×108)2E_k = (2 - 1) \times (1.6 \times 10^{-27}) \times (3.0 \times 10^8)^2

Ek=1.50×1010 JE_k = 1.50 \times 10^{-10} \text{ J}

Marking Criteria
DescriptorMarks

Demonstrates correct working using the relativistic kinetic energy formula, e.g., Ek=(γ1)mc2=(21)×(1.67×1027)×(3.0×108)2E_k = (\gamma - 1)mc^2 = (2 - 1) \times (1.67 \times 10^{-27}) \times (3.0 \times 10^8)^2

1

Calculates the correct kinetic energy, 1.50×1010 J1.50 \times 10^{-10} \text{ J}

1
Q33
2023
NESA
9 marks
Q33
9 marks

Consider the following statement.

"The interaction of subatomic particles with fields, as well as with other types of particles and matter, has increased our understanding of processes that occur in the physical world and of the properties of the subatomic particles themselves."

Justify this statement with reference to observations that have been made and experiments that scientists have carried out.

Reveal Answer
Marking Criteria
DescriptorMarks

Provides a reasoned detailed justification for the statement with explanation referencing at least TWO observations and at least TWO experiments

9

Provides a justification for the statement with explanation referencing at least TWO observations and at least TWO experiments

8

The student response meets all criteria of the 6-mark band, and additionally meets the majority of criteria in the 8-mark band.

7

Provides a justification for the statement with an explanation of an observation/experiment of particle–field interactions and an observation/experiment of particle–particle interactions

6

The student response meets all criteria of the 4-mark band, and additionally meets the majority of criteria in the 6-mark band.

5

Provides details of TWO experiments or observations and how they relate to the statement

4

The student response meets all criteria of the 2-mark band, and additionally meets the majority of criteria in the 4-mark band.

3
  • Relates the statement to an experiment or observation
    OR
  • Describes relevant experiments or observations
2

Provides some relevant information

1

None of the above

0
Q14
2023
VCAA
6 marks
Q14

Neutrons are subatomic particles and, like electrons, they can exhibit both particle-like and wave-like behaviour. Ignore any relativistic effects.

A beam of neutrons that can be used for scientific experiments is produced by a nuclear research reactor.

The mass of a neutron is 1.67×1027 kg1.67 \times 10^{-27} \text{ kg}.

The de Broglie wavelength of the neutrons produced by the nuclear reactor is 3.02×1010 m3.02 \times 10^{-10} \text{ m}.

Q14a
2 marks

Calculate the speed of the neutrons.

Reveal Answer

λd=hmv\lambda_d = \frac{h}{mv}

3.02×1010=6.63×1034(1.67×1027)×v3.02 \times 10^{-10} = \frac{6.63 \times 10^{-34}}{(1.67 \times 10^{-27}) \times v}

v=1.3×103 m s1v = 1.3 \times 10^3 \text{ m s}^{-1}

Marking Criteria
DescriptorMarks

Shows correct substitution into the de Broglie wavelength formula, e.g. 3.02×1010=6.63×1034(1.67×1027)×v3.02 \times 10^{-10} = \frac{6.63 \times 10^{-34}}{(1.67 \times 10^{-27}) \times v}

1

Calculates the correct speed of 1.3×103 m s11.3 \times 10^3 \text{ m s}^{-1}

1
Q14b
2 marks

The neutron beam is sent through a crystal with an interatomic spacing of 3.62×1010 m3.62 \times 10^{-10} \text{ m}.

Would you expect to observe a diffraction pattern? Justify your answer.

Reveal Answer

The ratio of λw=0.83\frac{\lambda}{w} = 0.83. As this ratio is close to 1, we would expect to be able to observe a diffraction pattern.

Marking Criteria
DescriptorMarks

Calculates the ratio λw\frac{\lambda}{w} as 0.830.83 and identifies that it is close to 1

1

Concludes that a diffraction pattern is expected to be observed

1
Q14c
2 marks

Consider an electron beam with the same de Broglie wavelength as the neutron beam, 3.02×1010 m3.02 \times 10^{-10} \text{ m}.

Which will have the greater speed: an electron in the electron beam or a neutron in the neutron beam? Justify your answer.

Reveal Answer

The electron will have the greater speed.

λd=hmv\lambda_d = \frac{h}{mv}

The product of mass and velocity must remain constant for wavelength to remain constant, so if mass decreases, velocity must increase.

Marking Criteria
DescriptorMarks

States that the electron will have the greater speed

1

Provides correct reasoning based on the de Broglie wavelength formula, explaining that for a constant wavelength, the product of mass and velocity must remain constant, so a smaller mass requires a greater velocity

1
Q8
2022
QCAA
Paper 1
1 mark
Q8
1 mark

Determine the wavelength of an electromagnetic wave with an energy of 2.4×10232.4 \times 10^{-23} J.

A

7.2×10157.2 \times 10^{-15} m

B

2.8×10112.8 \times 10^{-11} m

C

8.3×1038.3 \times 10^{-3} m

D

1.2×1021.2 \times 10^2 m

Reveal Answer
A

7.2×10157.2 \times 10^{-15} m

This incorrect value results from simply multiplying the energy by the speed of light (E×cE \times c), ignoring Planck's constant and the correct formula.

B

2.8×10112.8 \times 10^{-11} m

This answer is obtained by dividing Planck's constant by the energy (h/Eh/E) but neglecting to multiply by the speed of light (cc).

C

8.3×1038.3 \times 10^{-3} m

Correct Answer

Using the relationship λ=hcE\lambda = \frac{hc}{E}, substitute Planck's constant (h6.626×1034h \approx 6.626 \times 10^{-34} J\cdots) and the speed of light (c3.00×108c \approx 3.00 \times 10^8 m/s) to find λ8.3×103\lambda \approx 8.3 \times 10^{-3} m.

D

1.2×1021.2 \times 10^2 m

This value represents the wavenumber (1/λ1201/\lambda \approx 120 m1^{-1}), which is the reciprocal of the wavelength rather than the wavelength itself.

Q17
2021
VCAA
1 mark
Q17
1 mark

Which one of the following is closest to the de Broglie wavelength of a 663 kg663 \text{ kg} motor car moving at 10 m s110 \text{ m s}^{-1}?

A

1037 m10^{-37} \text{ m}

B

1036 m10^{-36} \text{ m}

C

1035 m10^{-35} \text{ m}

D

1034 m10^{-34} \text{ m}

Reveal Answer
A

1037 m10^{-37} \text{ m}

Correct Answer

Correct. Using the de Broglie wavelength formula λ=hmv\lambda = \frac{h}{mv}, we calculate λ=6.63×1034 J s663 kg×10 m s1=1037 m\lambda = \frac{6.63 \times 10^{-34} \text{ J s}}{663 \text{ kg} \times 10 \text{ m s}^{-1}} = 10^{-37} \text{ m}.

B

1036 m10^{-36} \text{ m}

Incorrect. This result is off by a factor of 10, which would occur if the velocity was 1 m s11 \text{ m s}^{-1} instead of 10 m s110 \text{ m s}^{-1}.

C

1035 m10^{-35} \text{ m}

Incorrect. This answer is off by a factor of 100, likely due to a miscalculation of the momentum denominator mv=6630 kg m s1mv = 6630 \text{ kg m s}^{-1}.

D

1034 m10^{-34} \text{ m}

Incorrect. This is approximately the value of Planck's constant (6.63×1034 J s6.63 \times 10^{-34} \text{ J s}), which means the momentum mvmv was incorrectly treated as 1 kg m s11 \text{ kg m s}^{-1}.

Q32
2024
NESA
8 marks
Q32
8 marks

Many scientists have performed experiments to explore the interaction of light and matter.

Analyse how evidence from at least THREE such experiments has contributed to our understanding of physics.

Reveal Answer

Answers could include:

Reference to:

  • Black body radiation experiments and the development of quantum physics
  • Photoelectric experiments and the development of quantum physics
  • Spectroscopy experiments and the development of astrophysics and the atomic model
  • Polarisation experiments and the development of the wave nature of light
  • Interference and diffraction and the development of the wave model of light
  • Cosmic gamma rays and the development of theory of special relativity and/or the standard model.
Marking Criteria
DescriptorMarks
  • Provides a detailed analysis using evidence from at least THREE experiments investigating the interaction of light and matter
  • Provides a clear link between experimental evidence and greater understanding of physics
8
  • Provides analysis using evidence from experiments investigating the interaction of light and matter
  • Provides a link between experimental evidence and greater understanding of physics
7

The student response meets all criteria of the 5-mark band, and additionally meets the majority of criteria in the 7-mark band.

6
  • Provides evidence from experiments investigating the interaction of light and matter
  • Relates evidence to a greater understanding of physics
5

The student response meets all criteria of the 3-mark band, and additionally meets the majority of criteria in the 5-mark band.

4

Provides some information about evidence from an experiment AND/OR a link to physics

3

The student response meets all criteria of the 1-mark band, and additionally meets the majority of criteria in the 3-mark band.

2

Provides some relevant information

1

None of the above

0
Q19
2021
QCAA
Paper 1
1 mark
Q19
1 mark

A spaceship with a velocity of 9.0×1079.0 \times 10^7 m s1^{-1} is measured to be 125 m in length by an observer at rest.
Calculate the length of the spaceship as measured by somebody on board the spaceship.

A

119 m

B

131 m

C

137 m

D

178 m

Reveal Answer
A

119 m

This calculation incorrectly treats the given 125 m as the proper length (L0L_0) and solves for the contracted length. Since the observer on the spaceship is at rest relative to it, they measure the proper length, which must be longer than the contracted length measured by the outside observer.

B

131 m

Correct Answer

The observer on board measures the proper length (L0L_0). Using the length contraction formula L=L01v2c2L = L_0 \sqrt{1 - \frac{v^2}{c^2}}, with L=125L=125 m and v=0.3cv=0.3c, we rearrange to find L0=12510.09131L_0 = \frac{125}{\sqrt{1 - 0.09}} \approx 131 m.

C

137 m

This answer results from omitting the square root in the Lorentz factor calculation, dividing 125 by (10.32)(1 - 0.3^2) instead of 10.32\sqrt{1 - 0.3^2}.

D

178 m

This value is incorrect and does not align with the standard length contraction formula. It likely results from a calculation error or misapplication of the Lorentz factor γ\gamma.

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=Δmc2E = \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.

Q6
2021
QCAA
Paper 1
1 mark
Q6
1 mark

Electromagnetic radiation is

A

extremely high-frequency radiation emitted from the nucleus of some radionuclides.

B

the emission of energy as waves or particles, especially high-energy particles, that causes ionisation.

C

a wave of energy produced by an oscillating electric charge, resulting in mutually perpendicular electric and magnetic fields.

D

radiant energy consisting of synchronised oscillations of electric and magnetic fields, or electromagnetic waves, propagated at the speed of light in a vacuum.

Reveal Answer
A

extremely high-frequency radiation emitted from the nucleus of some radionuclides.

This describes gamma rays specifically, which are just one type of high-energy electromagnetic radiation, rather than the general definition for all electromagnetic radiation.

B

the emission of energy as waves or particles, especially high-energy particles, that causes ionisation.

This defines ionizing radiation, which includes high-energy particles (like alpha and beta particles) in addition to waves, and excludes non-ionizing electromagnetic radiation like radio waves and visible light.

C

a wave of energy produced by an oscillating electric charge, resulting in mutually perpendicular electric and magnetic fields.

While this correctly describes the mechanism of generation and the geometry of an electromagnetic wave, Option D provides the more complete and formal definition of the radiant energy itself and its propagation characteristics.

D

radiant energy consisting of synchronised oscillations of electric and magnetic fields, or electromagnetic waves, propagated at the speed of light in a vacuum.

Correct Answer

This is the comprehensive definition of electromagnetic radiation, characterizing it as radiant energy composed of synchronized oscillating electric and magnetic fields that propagate at the speed of light (cc) in a vacuum.

Q11
2022
QCAA
Paper 1
1 mark
Q11
1 mark

The maximum kinetic energy of an electron ejected from a metallic surface can be increased by

A

using a positively ionised metal.

B

using a metal with a larger work function.

C

increasing the intensity of the incident light.

D

decreasing the wavelength of the incident light.

Reveal Answer
A

using a positively ionised metal.

Using a positively ionised metal would increase the electrostatic attraction on the electrons, effectively increasing the energy required to escape and decreasing the kinetic energy.

B

using a metal with a larger work function.

According to the photoelectric equation Kmax=hνΦK_{max} = h\nu - \Phi, increasing the work function (Φ\Phi) reduces the remaining energy available for kinetic energy.

C

increasing the intensity of the incident light.

Increasing intensity increases the number of photons and ejected electrons (photocurrent), but does not change the energy of individual photons or the maximum kinetic energy of the electrons.

D

decreasing the wavelength of the incident light.

Correct Answer

Photon energy is inversely proportional to wavelength (E=hcλE = \frac{hc}{\lambda}). Decreasing the wavelength increases the incident photon energy, thereby increasing the maximum kinetic energy of the ejected electrons.

Q14
2023
QCAA
Paper 1
1 mark
Q14
1 mark

An electron is best described as a

A

lepton with a larger mass than a positron.

B

baryon with a smaller mass than a proton.

C

meson that experiences the strong nuclear force.

D

particle whose interactions can be mediated by photons.

Reveal Answer
A

lepton with a larger mass than a positron.

Incorrect. While an electron is a lepton, it has the exact same mass as its antiparticle, the positron.

B

baryon with a smaller mass than a proton.

Incorrect. An electron is a lepton, not a baryon; baryons are composite particles made of three quarks (e.g., protons).

C

meson that experiences the strong nuclear force.

Incorrect. Electrons are leptons and do not experience the strong nuclear force, unlike mesons which are hadrons.

D

particle whose interactions can be mediated by photons.

Correct Answer

Correct. Electrons are electrically charged particles, and their electromagnetic interactions are mediated by the exchange of photons.

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