NESA Physics Quantum Mechanical Nature of the Atom

15 sample questions with marking guides and sample answers · Avg. score: 66.7%

Q9
2021
QCAA
Paper 1
1 mark
Q9
1 mark

The Bohr atomic model describes an atom as

A

the smallest particle of any substance.

B

a small dense nucleus orbited by electrons.

C

electrons scattered throughout a sphere of positively charged fluid.

D

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

Reveal Answer
A

the smallest particle of any substance.

This describes John Dalton's early atomic theory, which viewed atoms as indivisible spheres, rather than the internal structure proposed by Bohr.

B

a small dense nucleus orbited by electrons.

This describes Rutherford's planetary model. While similar, it fails to include Bohr's key innovation: that electrons are restricted to specific, quantized orbits.

C

electrons scattered throughout a sphere of positively charged fluid.

This describes J.J. Thomson's "Plum Pudding" model, which was disproved by Rutherford before Bohr developed his model.

D

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

Correct Answer

The Bohr model refined the nuclear model by proposing that electrons move in specific, circular orbits (shells) with fixed, quantized energy levels around the nucleus.

Q8
2022
QCAA
Paper 1
1 mark
Q8
1 mark

Determine the wavelength of an electromagnetic wave with an energy of 2.4×10−232.4 \times 10^{-23} J.

A

7.2×10−157.2 \times 10^{-15} m

B

2.8×10−112.8 \times 10^{-11} m

C

8.3×10−38.3 \times 10^{-3} m

D

1.2×1021.2 \times 10^2 m

Reveal Answer
A

7.2×10−157.2 \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

2.8×10−112.8 \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

8.3×10−38.3 \times 10^{-3} m

Correct Answer

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

D

1.2×1021.2 \times 10^2 m

This value represents the wavenumber (1/λ≈1201/\lambda \approx 120 m−1^{-1}), which is the reciprocal of the wavelength rather than the wavelength itself.

Q10
2023
SCSA
4 marks
Q10
4 marks

Estimate the de Broglie wavelength for a standard men's basketball travelling at 10.0 m s−1^{-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=6.63×10−34(0.62)(10.0)=1.07×10−34 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

λ≈1.1×10−34 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
Q18
2020
VCAA
1 mark
Q18
1 mark

Quantised energy levels within atoms can best be explained by

A

electrons behaving as individual particles with different energies.

B

electrons behaving as waves, with each energy level representing a diffraction pattern.

C

protons behaving as waves, with only standing waves at particular wavelengths allowed.

D

electrons behaving as waves, with only standing waves at particular wavelengths allowed.

Reveal Answer
A

electrons behaving as individual particles with different energies.

Treating electrons purely as classical particles does not explain why only specific, discrete energy levels are allowed within an atom.

B

electrons behaving as waves, with each energy level representing a diffraction pattern.

While electrons do exhibit wave-like behavior, atomic energy levels represent standing waves, not diffraction patterns.

C

protons behaving as waves, with only standing waves at particular wavelengths allowed.

The quantised energy levels of an atom are determined by the wave-like behavior of electrons in the electron shells, not protons in the nucleus.

D

electrons behaving as waves, with only standing waves at particular wavelengths allowed.

Correct Answer

The wave nature of electrons means they form standing waves within the atom. Only specific wavelengths can form stable standing waves, which corresponds to the quantised energy levels.

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
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:

E=1.81 keV=1810 eV=1810(1.60×10−19)=2.896×10−16 J\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=hfE = hf and f=cλf = \frac{c}{\lambda},

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

Rearranging for λ\lambda:

λ=hcE=(6.63×10−34)(3.00×108)2.896×10−16=6.87×10−10 m\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
DescriptorMarks

Converts to joules

1

Substitutes cλ\frac{c}{\lambda} for ff

1

Rearranges for λ\lambda

1

Calculates answer

1
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 s−110 \text{ m s}^{-1}?

A

10−37 m10^{-37} \text{ m}

B

10−36 m10^{-36} \text{ m}

C

10−35 m10^{-35} \text{ m}

D

10−34 m10^{-34} \text{ m}

Reveal Answer
A

10−37 m10^{-37} \text{ m}

Correct Answer

Correct. Using the de Broglie wavelength formula λ=hmv\lambda = \frac{h}{mv}, we calculate λ=6.63×10−34 J s663 kg×10 m s−1=10−37 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

10−36 m10^{-36} \text{ m}

Incorrect. This result is off by a factor of 10, which would occur if the velocity was 1 m s−11 \text{ m s}^{-1} instead of 10 m s−110 \text{ m s}^{-1}.

C

10−35 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 s−1mv = 6630 \text{ kg m s}^{-1}.

D

10−34 m10^{-34} \text{ m}

Incorrect. This is approximately the value of Planck's constant (6.63×10−34 J s6.63 \times 10^{-34} \text{ J s}), which means the momentum mvmv was incorrectly treated as 1 kg m s−11 \text{ kg m s}^{-1}.

Q20
2023
VCAA
1 mark
Q20
1 mark

Heisenberg's uncertainty principle can be used to explain the results of single-slit diffraction experiments for electrons.

This is because Heisenberg's uncertainty principle

A

fixes the size range of the slit to be used.

B

states that each electron's exact position was predictable after passing through the slit.

C

states that each electron's actual position after passing though the slit was only known within a wide range.

D

states that if the electron's momentum was known, its position after passing through the slit was also known.

Reveal Answer
A

fixes the size range of the slit to be used.

The uncertainty principle relates the uncertainties in position and momentum; it does not dictate or fix the physical size of the slit used in an experiment.

B

states that each electron's exact position was predictable after passing through the slit.

The uncertainty principle fundamentally denies the ability to predict an exact position and momentum simultaneously, meaning the exact path of the electron is unpredictable.

C

states that each electron's actual position after passing though the slit was only known within a wide range.

Correct Answer

By restricting the electron's position at the slit (small Δx\Delta x), its transverse momentum becomes highly uncertain (large Δp\Delta p), causing the electron to spread out and hit the screen over a wide range of positions.

D

states that if the electron's momentum was known, its position after passing through the slit was also known.

This contradicts the uncertainty principle, which states that the more precisely the momentum is known, the less precisely the position can be known.

Q21
2023
QCAA
Paper 1
4 marks
Q21
4 marks

Describe how the atomic model proposed by Bohr addresses the limitation of Rutherford’s model.

Reveal Answer

The Rutherford model provided evidence for the presence of electrons around the nucleus of an atom, while the Bohr atomic model described how electrons orbit the nucleus.

However, Rutherford's model was limited, because it couldn't account for the stability of atoms, as electrons orbiting the nucleus would gradually lose energy and spiral into the nucleus. The Bohr model addressed this by explaining the quantised nature of these orbits and how electrons within the same orbit possess the same discrete quantity of energy.

Marking Criteria
DescriptorMarks

Describes Rutherford's atomic model

1

Describes Bohr's atomic model

1

Describes a limitation of Rutherford's atomic model

1

Describes significance of quantised energy levels

1
Q26
2023
QCAA
Paper 1
4 marks
Q26
4 marks

Calculate the energy (in electron volts) of a photon with a wavelength of 405 nm. Show your working.
Energy = ______ eV (three significant figures)

Reveal Answer

f=cλf = \frac{c}{\lambda}
=3×108405×10−9= \frac{3 \times 10^8}{405 \times 10^{-9}}
=7.41×1014 Hz= 7.41 \times 10^{14} \text{ Hz}

E=hfE = hf
=6.626×10−34×7.41×1014= 6.626 \times 10^{-34} \times 7.41 \times 10^{14}
=4.91×10−19 J= 4.91 \times 10^{-19} \text{ J}
=3.07 eV= 3.07 \text{ eV}

Energy =3.07 eV= 3.07 \text{ eV} (to three significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to conversion of wavelength to frequency

1

Provides appropriate mathematical reasoning

1

Calculates the energy of photon

1

Converts energy into eV

1
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 1.67×10−27 kg1.67 \times 10^{-27} \text{ kg}.

The de Broglie wavelength of the neutrons produced by the nuclear reactor is 3.02×10−10 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}

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

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

Marking Criteria
DescriptorMarks

Shows correct substitution into the de Broglie wavelength formula, e.g. 3.02×10−10=6.63×10−34(1.67×10−27)×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 s−11.3 \times 10^3 \text{ m s}^{-1}

1
Q14b
2 marks

The neutron beam is sent through a crystal with an interatomic spacing of 3.62×10−10 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, 3.02×10−10 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
Q17
2022
VCAA
1 mark
Q17
1 mark

Gamma radiation is often used to treat cancerous tumours. The energy of a gamma photon emitted by radioactive cobalt-60 is 1.33 MeV.

Which one of the following is closest to the frequency of the gamma radiation?

A

1.33×106 Hz1.33 \times 10^6 \text{ Hz}

B

3.21×1020 Hz3.21 \times 10^{20} \text{ Hz}

C

3.21×1021 Hz3.21 \times 10^{21} \text{ Hz}

D

2.01×1039 Hz2.01 \times 10^{39} \text{ Hz}

Reveal Answer
A

1.33×106 Hz1.33 \times 10^6 \text{ Hz}

This is incorrect because it simply takes the numerical value of the energy in eV (1.33×1061.33 \times 10^6) and assigns it the unit of frequency (Hz) without applying Planck's equation.

B

3.21×1020 Hz3.21 \times 10^{20} \text{ Hz}

Correct Answer

This is correct. Using Planck's equation E=hfE = hf, the frequency is f=Ehf = \frac{E}{h}. Using the energy in eV (1.33×106 eV1.33 \times 10^6 \text{ eV}) and Planck's constant in eV s (4.14×10−15 eV s4.14 \times 10^{-15} \text{ eV s}), we get f=3.21×1020 Hzf = 3.21 \times 10^{20} \text{ Hz}.

C

3.21×1021 Hz3.21 \times 10^{21} \text{ Hz}

This is incorrect. The value is off by a factor of 10, which likely results from a simple arithmetic or unit conversion error when calculating f=Ehf = \frac{E}{h}.

D

2.01×1039 Hz2.01 \times 10^{39} \text{ Hz}

This is incorrect because it results from dividing the energy in eV (1.33×106 eV1.33 \times 10^6 \text{ eV}) by Planck's constant in Joules seconds (6.63×10−34 J s6.63 \times 10^{-34} \text{ J s}). Units must be consistent (either both in eV or both in Joules) before dividing.

Q18
2021
VCAA
5 marks
Q18

Scientists are conducting experiments to compare the circular diffraction patterns formed by X-ray photons and electrons when they pass through small circular apertures. The X-ray photons have an energy of 100 eV100 \text{ eV} and pass through an aperture of diameter 1.24 μm1.24 \text{ } \mu\text{m}. The electrons are moving at 5.0×105 m s−15.0 \times 10^5 \text{ m s}^{-1}.

Q18a
1 mark

Show that the de Broglie wavelength of the electrons is equal to 1.46×10−9 m1.46 \times 10^{-9} \text{ m}.

Reveal Answer

λ=hmv\lambda = \frac{h}{mv}
λ=6.63×10−349.1×10−31×5.0×105\lambda = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.0 \times 10^5}
λ=1.46×10−9 m\lambda = 1.46 \times 10^{-9} \text{ m}

Marking Criteria
DescriptorMarks

Calculates the correct de Broglie wavelength of the electrons (1.46×10−9 m1.46 \times 10^{-9} \text{ m}) by substituting correct values into the de Broglie equation

1
Q18b
4 marks

The scientists want an aperture for the electrons that forms diffraction patterns with the same spacing as the diffraction patterns formed by the X-ray photons.

Calculate the diameter of the aperture that the scientists should choose. Show your working.

Reveal Answer

The width of the diffraction pattern can be found from the λw\frac{\lambda}{w} ratio. The wavelength of the X-rays is found by:
E=hcλE = \frac{hc}{\lambda}
100=4.14×10−15×3.0×108λ100 = \frac{4.14 \times 10^{-15} \times 3.0 \times 10^8}{\lambda}
λ=1.24×10−8 m\lambda = 1.24 \times 10^{-8} \text{ m}
This gives a ratio of:
λw=1.24×10−81.24×10−6\frac{\lambda}{w} = \frac{1.24 \times 10^{-8}}{1.24 \times 10^{-6}}
λw=1.00×10−2\frac{\lambda}{w} = 1.00 \times 10^{-2}
The electrons, with a de Broglie wavelength of 1.46×10−9 m1.46 \times 10^{-9} \text{ m}, will also have to have the same ratio of λw\frac{\lambda}{w}.
1.46×10−9w=1.00×10−2\frac{1.46 \times 10^{-9}}{w} = 1.00 \times 10^{-2}
w=1.46×10−7 mw = 1.46 \times 10^{-7} \text{ m}

Marking Criteria
DescriptorMarks

Calculates the wavelength of the X-rays (1.24×10−8 m1.24 \times 10^{-8} \text{ m})

1

Calculates the ratio of wavelength to slit width for the X-rays (1.00×10−21.00 \times 10^{-2})

1

Equates the ratio for the electrons to the ratio for the X-rays

1

Calculates the correct slit width for the electrons (1.46×10−7 m1.46 \times 10^{-7} \text{ m})

1
Q18
2021
VCAA
1 mark
Q18
1 mark

A monochromatic light source is emitting green light with a wavelength of 550 nm550 \text{ nm}. The light source emits 2.8×10162.8 \times 10^{16} photons every second.

Which one of the following is closest to the power of the light source?

A

1.0×10−2 W1.0 \times 10^{-2} \text{ W}

B

3.3×10−11 W3.3 \times 10^{-11} \text{ W}

C

2.1×109 W2.1 \times 10^9 \text{ W}

D

6.3×1016 W6.3 \times 10^{16} \text{ W}

Reveal Answer
A

1.0×10−2 W1.0 \times 10^{-2} \text{ W}

Correct Answer

Power is the total energy emitted per second. The energy of a single photon is E=hcλE = \frac{hc}{\lambda}. Multiplying this by the number of photons per second (nn) gives P=nhcλ=(2.8×1016)(6.63×10−34)(3×108)550×10−9≈1.0×10−2 WP = n \frac{hc}{\lambda} = (2.8 \times 10^{16}) \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{550 \times 10^{-9}} \approx 1.0 \times 10^{-2} \text{ W}.

B

3.3×10−11 W3.3 \times 10^{-11} \text{ W}

This incorrect value results from omitting the speed of light (cc) from the photon energy equation, calculating P=nhλP = n \frac{h}{\lambda} instead of P=nhcλP = n \frac{hc}{\lambda}.

C

2.1×109 W2.1 \times 10^9 \text{ W}

This incorrect value results from multiplying the number of photons by the wavelength and dividing by Planck's constant and the speed of light, which does not represent the physical formula for power.

D

6.3×1016 W6.3 \times 10^{16} \text{ W}

This incorrect value results from improperly calculating the energy of the photons or misapplying the power formula. Power must be calculated using P=nhcλP = n \frac{hc}{\lambda}.

Q14
2022
VCAA
1 mark
Q14
1 mark

Which one of the following best provides evidence of electrons behaving as waves?

A

photoelectric effect

B

atomic emission spectra

C

atomic absorption spectra

D

diffraction of electrons through a crystal

Reveal Answer
A

photoelectric effect

Incorrect. The photoelectric effect demonstrates that light behaves as a particle (a photon), rather than showing the wave-like nature of electrons.

B

atomic emission spectra

Incorrect. Atomic emission spectra provide evidence for discrete, quantized energy levels within an atom, rather than the wave-like behavior of free electrons.

C

atomic absorption spectra

Incorrect. Like emission spectra, atomic absorption spectra demonstrate that electrons transition between quantized energy states, which does not directly illustrate wave behavior.

D

diffraction of electrons through a crystal

Correct Answer

Correct. Diffraction is a characteristic property of waves. The ability of electrons to diffract through a crystal lattice directly proves their wave-like nature, confirming the de Broglie relationship λ=hp\lambda = \frac{h}{p}.

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