Module 8: From the Universe to the Atom — Worked Solutions (HSC Physics)
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Worked examples for HSC Physics Module 8: From the Universe to the Atom. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Physics — Module 8: From the Universe to the Atom. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Useful constants: Planck's constant $h = 6.63 \times 10^{-34}\ \text{J s}$, speed of light $c = 3.0 \times 10^8\ \text{m s}^{-1}$, and $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$.
Example 1 — Atomic energy levels
Question
A hydrogen atom emits a photon when an electron transitions from an energy level of $-1.51\ \text{eV}$ to a level of $-3.40\ \text{eV}$. Using $h = 6.63 \times 10^{-34}\ \text{J s}$, $c = 3.0 \times 10^8\ \text{m s}^{-1}$ and $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$, calculate the wavelength of the emitted photon.
Solution
The photon energy equals the drop in level energy: $\Delta E = E_{high} - E_{low}$.
$$\Delta E = (-1.51) - (-3.40) = 1.89\ \text{eV}$$
Convert to joules: $\Delta E = 1.89 \times 1.6\times10^{-19} = 3.024\times10^{-19}\ \text{J}$.
Now use $E = \dfrac{hc}{\lambda}$, rearranged for wavelength:
$$\lambda = \frac{hc}{\Delta E} = \frac{(6.63\times10^{-34})(3.0\times10^8)}{3.024\times10^{-19}} = 6.58\times10^{-7}\ \text{m}$$
So $\lambda \approx 6.58\times10^{-7}\ \text{m}$ — red light, as expected for this Balmer transition. Take the difference of the levels first, then convert; don't convert each level separately.
When an electron falls from a higher to a lower energy level, the atom releases the energy difference as a single photon. So the first step is the energy gap:
$$\Delta E = E_{high} - E_{low} = (-1.51) - (-3.40) = 1.89\ \text{eV}$$
(It's positive because the electron loses energy, and that energy goes into the photon.) We then convert to joules so it works with our constants: $\Delta E = 1.89 \times 1.6\times10^{-19} = 3.024\times10^{-19}\ \text{J}$.
Each photon's energy is linked to its wavelength by $E = \dfrac{hc}{\lambda}$, so rearranging for $\lambda$:
$$\lambda = \frac{hc}{\Delta E} = \frac{(6.63\times10^{-34})(3.0\times10^8)}{3.024\times10^{-19}} = 6.58\times10^{-7}\ \text{m}$$
That's about $658\ \text{nm}$, which is red light — and indeed this is one of the visible hydrogen emission lines. The key idea is that quantised energy levels produce a discrete spectrum, one wavelength per allowed transition.
Energy gap, convert, then wavelength.
- $\Delta E = (-1.51) - (-3.40) = 1.89\ \text{eV}$
- $\Delta E = 1.89 \times 1.6\times10^{-19} = 3.024\times10^{-19}\ \text{J}$
- $\lambda = \dfrac{hc}{\Delta E} = \dfrac{(6.63\times10^{-34})(3.0\times10^8)}{3.024\times10^{-19}}$
- $\lambda = 6.58\times10^{-7}\ \text{m}$
Where the marks go
- 1 mark: Correct energy difference $\Delta E = 1.89\ \text{eV}$ converted to $3.024\times10^{-19}\ \text{J}$
- 1 mark: Uses $\lambda = \dfrac{hc}{\Delta E}$ with correct substitution
- 1 mark: Correct wavelength $\lambda = 6.58\times10^{-7}\ \text{m}$ with units
Key idea
A photon carries the energy gap between levels: $\Delta E = \dfrac{hc}{\lambda}$, so $\lambda = \dfrac{hc}{\Delta E}$ — quantised levels give a discrete emission spectrum.
Example 2 — Nuclear mass defect and binding energy
Question
A helium-4 nucleus is made of 2 protons and 2 neutrons. The mass of a proton is $1.6726 \times 10^{-27}\ \text{kg}$, the mass of a neutron is $1.6749 \times 10^{-27}\ \text{kg}$, and the measured mass of the helium-4 nucleus is $6.6447 \times 10^{-27}\ \text{kg}$. Using $c = 3.0 \times 10^8\ \text{m s}^{-1}$, calculate the mass defect and the binding energy of the nucleus.
Solution
First add up the masses of the separate nucleons:
$$2(1.6726\times10^{-27}) + 2(1.6749\times10^{-27}) = 6.6950\times10^{-27}\ \text{kg}$$
The mass defect is the difference between this and the measured nuclear mass:
$$\Delta m = 6.6950\times10^{-27} - 6.6447\times10^{-27} = 5.03\times10^{-29}\ \text{kg}$$
Convert that "missing" mass to energy with $E = \Delta m c^2$:
$$E = (5.03\times10^{-29})(3.0\times10^8)^2 = 4.53\times10^{-12}\ \text{J}$$
Mass defect $5.03\times10^{-29}\ \text{kg}$, binding energy $4.53\times10^{-12}\ \text{J}$. The defect is tiny — keep all your significant figures until the final step.
A nucleus weighs slightly less than the sum of its individual protons and neutrons. That missing mass is the mass defect, and it has been converted into the energy that binds the nucleus together.
Start by totalling the nucleon masses:
$$2(1.6726\times10^{-27}) + 2(1.6749\times10^{-27}) = 6.6950\times10^{-27}\ \text{kg}$$
Subtract the actual nuclear mass to find what's "missing":
$$\Delta m = 6.6950\times10^{-27} - 6.6447\times10^{-27} = 5.03\times10^{-29}\ \text{kg}$$
Einstein's $E = \Delta m c^2$ tells us how much energy that mass is worth — and that's the binding energy, the energy you'd need to pull the nucleus apart:
$$E = (5.03\times10^{-29})(3.0\times10^8)^2 = 4.53\times10^{-12}\ \text{J}$$
So the mass defect is $5.03\times10^{-29}\ \text{kg}$ and the binding energy is $4.53\times10^{-12}\ \text{J}$. The deeper idea is that mass and energy are interchangeable — the strong nuclear force locks the nucleons together, and that "lost" mass is the energy price of staying bound.
Sum nucleons, subtract, then $E = \Delta m c^2$.
- Total nucleon mass $= 2(1.6726\times10^{-27}) + 2(1.6749\times10^{-27}) = 6.6950\times10^{-27}\ \text{kg}$
- $\Delta m = 6.6950\times10^{-27} - 6.6447\times10^{-27} = 5.03\times10^{-29}\ \text{kg}$
- $E = \Delta m c^2 = (5.03\times10^{-29})(3.0\times10^8)^2$
- $E = 4.53\times10^{-12}\ \text{J}$
Where the marks go
- 1 mark: Correct total nucleon mass $6.6950\times10^{-27}\ \text{kg}$
- 1 mark: Correct mass defect $\Delta m = 5.03\times10^{-29}\ \text{kg}$
- 1 mark: Applies $E = \Delta m c^2$ with correct substitution
- 1 mark: Correct binding energy $E = 4.53\times10^{-12}\ \text{J}$ with units
Key idea
The mass defect $\Delta m$ is the difference between the sum of nucleon masses and the measured nuclear mass; the binding energy is $E = \Delta m c^2$ — the energy equivalent of the missing mass.
Frequently asked questions
Step-by-step solutions to exam-style questions on From the Universe to the Atom in HSC Physics, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Physics syllabus for From the Universe to the Atom, using the methods and notation expected in the exam.