QCAA Physics Alternative Sequence Quantum theory

15 sample questions with marking guides and sample answers

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
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
DescriptorMarks
  • 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
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.

A positively ionised metal would exert a stronger attractive force on the electrons, effectively increasing the energy required to remove them and decreasing their maximum kinetic energy.

B

using a metal with a larger work function.

According to the photoelectric equation Kmax=hfΦK_{\text{max}} = hf - \Phi, using a metal with a larger work function (Φ\Phi) would decrease the maximum kinetic energy of the ejected electrons.

C

increasing the intensity of the incident light.

Increasing the intensity of the incident light increases the number of ejected electrons per second, but it 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

Decreasing the wavelength of the incident light increases the energy of the incident photons (E=hcλE = \frac{hc}{\lambda}), which directly increases the maximum kinetic energy of the ejected electrons according to Kmax=hcλΦK_{\text{max}} = \frac{hc}{\lambda} - \Phi.

Q27
2025
NESA
3 marks
Q27
3 marks

Outline TWO ways in which Schrödinger's model of electron behaviour is different from electron behaviour in the atomic models of Rutherford and Bohr.

Reveal Answer

In contrast to Bohr's idea of fixed orbits, Schrdinger described orbitals as probabilities of electrons as being in particular locations.

Unlike Rutherford's model in which electrons were imagined as particles orbiting the nucleus, Schrdinger described the electrons as waves, based on the work of de Broglie.

Marking Criteria
DescriptorMarks

Outlines TWO ways in which Schrödinger's model differs from those of Rutherford and Bohr

3
  • Outlines ONE way in which Schrödinger's model differs from those of Rutherford and Bohr
    OR
  • Outlines TWO features of Schrödinger's model
2

Provides some relevant information

1

None of the above

0
Q8
2022
QCAA
Paper 1
1 mark
Q8
1 mark

Determine the wavelength of an electromagnetic wave with an energy of 2.4×1023 J2.4 \times 10^{-23} \text{ J}.

A

7.2×1015 m7.2 \times 10^{-15} \text{ m}

B

2.8×1011 m2.8 \times 10^{-11} \text{ m}

C

8.3×103 m8.3 \times 10^{-3} \text{ m}

D

1.2×102 m1.2 \times 10^2 \text{ m}

Reveal Answer
A

7.2×1015 m7.2 \times 10^{-15} \text{ m}

This incorrect value is obtained by multiplying the energy by the speed of light (EcE \cdot c), rather than using the correct wave energy formula.

B

2.8×1011 m2.8 \times 10^{-11} \text{ m}

This incorrect value is obtained by dividing Planck's constant by the energy (h/Eh/E), which calculates the inverse of the frequency, not the wavelength.

C

8.3×103 m8.3 \times 10^{-3} \text{ m}

Correct Answer

Using the equation E=hcλE = \frac{hc}{\lambda}, we can rearrange to find λ=hcE\lambda = \frac{hc}{E}. Plugging in the constants gives λ=(6.626×1034 Js)(3.0×108 m/s)2.4×1023 J8.3×103 m\lambda = \frac{(6.626 \times 10^{-34} \text{ J}\cdot\text{s})(3.0 \times 10^8 \text{ m/s})}{2.4 \times 10^{-23} \text{ J}} \approx 8.3 \times 10^{-3} \text{ m}.

D

1.2×102 m1.2 \times 10^2 \text{ m}

This incorrect value is the inverse of the wavelength (1/λ1/\lambda), obtained by calculating Ehc\frac{E}{hc} instead of hcE\frac{hc}{E}.

Q7
2020
QCAA
Paper 1
1 mark
Q7
1 mark

A quantum of any form of electromagnetic radiation is also known as

A

a photon.

B

an X-ray.

C

a positron.

D

an electron.

Reveal Answer
A

a photon.

Correct Answer

A photon is the fundamental particle, or quantum, of all electromagnetic radiation, carrying energy proportional to its frequency.

B

an X-ray.

An X-ray is a specific type of high-energy electromagnetic radiation, not the general term for a discrete packet or quantum of radiation.

C

a positron.

A positron is the antimatter counterpart to an electron, having mass and a positive charge, whereas a quantum of electromagnetic radiation is a massless particle.

D

an electron.

An electron is a fundamental subatomic particle with mass and a negative charge, not a massless quantum of electromagnetic energy.

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
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
Q5
2024
QCAA
Paper 2
3 marks
Q5
3 marks

Explain the significance of the threshold frequency when incident light with a range of frequencies shines on a metal.

Reveal Answer

Threshold frequency is the minimum frequency of a photon required to eject a photoelectron from the surface of a metal. In this case, photons with frequencies equal to or greater than the threshold frequency will eject photoelectrons from the surface of the metal, while photons with a lower frequency will scatter off the metal's surface.

Marking Criteria
DescriptorMarks

describes threshold frequency as the minimum frequency of an incident photon required to eject a photoelectron

1

explains that photons with frequencies greater than the threshold frequency will have enough energy to eject a photoelectron

1

explains that photons with frequencies less than the threshold frequency will be scattered

1
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 classic wave phenomenon where overlapping waves constructively and destructively combine, demonstrating the wave nature of light.

B

equal width of the slits.

While slit width affects the intensity and diffraction envelope of the pattern, the actual phenomenon of interference is fundamentally caused by wave superposition, not the physical dimensions of the slits.

C

discrete packets of photons.

Describing light as discrete packets of photons refers to its particle nature, which explains phenomena like the photoelectric effect rather than classical wave interference.

D

distance from the slits to the screen.

The distance to the screen affects the spacing between the interference fringes (given by y=λLdy = \frac{\lambda L}{d}), but it does not explain the fundamental physical cause of the interference pattern itself.

Q15
2022
VCAA
1 mark
Q15
1 mark

Which one of the following best provides evidence of light behaving as a particle?

A

photoelectric effect

B

white light passing through a prism

C

diffraction of light through a single slit

D

interference of light passing through a double slit

Reveal Answer
A

photoelectric effect

Correct Answer

The photoelectric effect demonstrates that light energy is absorbed and emitted in discrete quantized packets called photons, providing direct evidence of its particle nature.

B

white light passing through a prism

White light passing through a prism demonstrates dispersion and refraction, which are phenomena explained by the wave nature of light and its varying wavelengths.

C

diffraction of light through a single slit

Diffraction is the bending of light as it passes through a slit, which is a classic characteristic of waves rather than particles.

D

interference of light passing through a double slit

Interference patterns are created by the superposition of overlapping waves, providing foundational evidence for the wave model of light.

Q6
2024
NESA
1 mark
Q6
1 mark

The photoelectric effect is mathematically modelled by the following relationship:

Kmax=hfϕK_{\text{max}} = hf - \phi

In this model, the symbol ϕ\phi represents the amount of energy

A

supplied by a photon to an electron.

B

retained by an electron after being hit.

C

required to release an electron from a material.

D

left over after a collision of a photon with an electron.

Reveal Answer
A

supplied by a photon to an electron.

Incorrect. The energy supplied by the incident photon is represented by the term hfhf, not ϕ\phi.

B

retained by an electron after being hit.

Incorrect. The energy retained by the electron as kinetic energy after being ejected is represented by KmaxK_{\text{max}}.

C

required to release an electron from a material.

Correct Answer

Correct. The symbol ϕ\phi represents the work function, which is the minimum energy required to overcome the attractive forces and release an electron from the material.

D

left over after a collision of a photon with an electron.

Incorrect. The energy left over after freeing the electron becomes its maximum kinetic energy, represented by KmaxK_{\text{max}}.

Q13
2020
QCAA
Paper 1
1 mark
Q13
1 mark

The Rutherford atomic model describes an atom

A

as the smallest particle of any substance.

B

with a small, dense nucleus surrounded by orbiting electrons.

C

consisting of electrons scattered throughout a sphere of positively charged fluid.

D

consisting of a small positive nucleus surrounded by negative electrons in set orbits of fixed energy.

Reveal Answer
A

as the smallest particle of any substance.

Incorrect. This describes John Dalton's early atomic theory. Rutherford's model focused on the internal structure of the atom rather than its indivisibility.

B

with a small, dense nucleus surrounded by orbiting electrons.

Correct Answer

Correct. Rutherford's gold foil experiment demonstrated that atoms consist mostly of empty space with a tiny, dense, positively charged nucleus at the center surrounded by electrons.

C

consisting of electrons scattered throughout a sphere of positively charged fluid.

Incorrect. This describes J.J. Thomson's plum pudding model. Rutherford's work specifically disproved this model by showing the positive charge is concentrated in a central nucleus.

D

consisting of a small positive nucleus surrounded by negative electrons in set orbits of fixed energy.

Incorrect. This describes the Bohr model. Rutherford proposed orbiting electrons, but it was Niels Bohr who introduced the concept of quantized orbits with fixed energy levels.

Q20
2020
QCAA
Paper 1
1 mark
Q20
1 mark

Calculate the frequency of light that would be required to eject a photoelectron at a velocity of 1.90×106 m s11.90 \times 10^6 \text{ m s}^{-1} from a metal plate with a work function of 4.73 eV.

A

1.14×10151.14 \times 10^{15} Hz

B

1.34×10151.34 \times 10^{15} Hz

C

2.48×10152.48 \times 10^{15} Hz

D

3.62×10153.62 \times 10^{15} Hz

Reveal Answer
A

1.14×10151.14 \times 10^{15} Hz

This is the threshold frequency, calculated using only the work function (Φ/h\Phi/h), which ignores the additional energy needed for the kinetic energy of the ejected electron.

B

1.34×10151.34 \times 10^{15} Hz

This frequency incorrectly corresponds to the difference between the kinetic energy and the work function, rather than their sum.

C

2.48×10152.48 \times 10^{15} Hz

This frequency corresponds only to the kinetic energy of the electron (K/hK/h), ignoring the energy required to overcome the metal's work function.

D

3.62×10153.62 \times 10^{15} Hz

Correct Answer

The total photon energy is the sum of the work function and the electron's kinetic energy (E=Φ+12mv2E = \Phi + \frac{1}{2}mv^2). Dividing this total energy by Planck's constant gives the correct frequency.

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