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QLDPhysicsSyllabus dot point

Topic 2: Quantum theory

Apply the photon model of light (E=hfE = hf), the photoelectric equation (Ek,max=hfϕE_{k,\max} = hf - \phi), and the Bohr model of atomic energy levels with transitions producing photons of energy ΔE=hf\Delta E = h f

A focused answer to the QCE Physics Unit 4 dot point on quantum theory. Planck's quantum hypothesis, Einstein's photon model, the photoelectric effect with work function and threshold frequency, the Bohr model of the hydrogen atom, and emission/absorption spectra.

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  1. What this dot point is asking
  2. Planck's quantum hypothesis
  3. The photoelectric effect
  4. The Bohr model of the hydrogen atom
  5. Photon emission and absorption
  6. Spectral series
  7. Emission and absorption spectra
  8. Ionisation
  9. Multi-electron atoms
  10. Examples in context
  11. Try this

What this dot point is asking

QCAA wants you to apply the photon model of light, the photoelectric equation, and the Bohr model of the hydrogen atom. The dot point integrates Topic 2 quantum theory and bridges to wave-particle duality (covered separately).

Planck's quantum hypothesis

In 1900, Max Planck postulated that the energy of electromagnetic radiation comes in discrete quanta of energy E=hfE = hf, with h=6.626×1034h = 6.626 \times 10^{-34} J s (Planck's constant). This was originally a mathematical trick to explain the spectrum of blackbody radiation (the ultraviolet catastrophe of classical physics). Planck did not initially believe the quanta were physical.

Einstein (1905) reinterpreted the quanta as physical particles of light: photons. Each photon has energy hfhf and momentum h/λh/\lambda. The photon model explained the photoelectric effect (next section) and ultimately the entire quantum nature of light.

The photoelectric effect

When light shines on a clean metal surface, electrons (photoelectrons) can be ejected. The phenomenon has four observations that the classical wave model cannot explain:

  1. Threshold frequency. Below a frequency f0f_0, no electrons are emitted, regardless of intensity.
  2. Instantaneous emission. Electrons emit essentially immediately (within nanoseconds), even at very low intensity, provided f>f0f > f_0.
  3. Intensity controls current, not maximum kinetic energy of each electron.
  4. Maximum kinetic energy depends linearly on frequency above threshold.

Einstein's photon model (1905) explained all four:

  • Each photon has energy E=hfE = hf.
  • A single photon interacts with a single electron.
  • The electron uses ϕ\phi (the work function) to escape and keeps the remainder as kinetic energy:

Ek,max=hfϕE_{k,\max} = hf - \phi

  • Below threshold (hf<ϕhf < \phi), no single photon has enough energy to eject an electron, regardless of intensity.
  • Intensity is the number of photons per second; it controls the current but not the energy of each electron.

Threshold frequency

f0=ϕhf_0 = \frac{\phi}{h}

This is the minimum frequency at which the photoelectric effect occurs.

Stopping voltage

The retarding voltage that stops all photoelectrons is the stopping voltage:

eV0=Ek,max=hfϕe V_0 = E_{k,\max} = hf - \phi

A plot of V0V_0 vs ff for a given metal is a straight line with gradient h/eh/e and xx-intercept f0f_0.

Cross-link: see the photoelectric calculator.

The Bohr model of the hydrogen atom

Bohr (1913) proposed that:

  1. Electrons in atoms occupy discrete, allowed energy levels EnE_n.
  2. Electrons in level nn do not radiate (despite their classical orbital acceleration).
  3. Electrons can transition between levels by emitting or absorbing photons.
  4. The angular momentum of the electron is quantised: L=nL = n \hbar where =h/(2π)\hbar = h / (2\pi).

For hydrogen, the energy levels are:

En=13.6 eVn2E_n = -\frac{13.6 \text{ eV}}{n^2}

for n=1,2,3,n = 1, 2, 3, \ldots.

  • E1=13.6E_1 = -13.6 eV (ground state).
  • E2=3.4E_2 = -3.4 eV.
  • E3=1.51E_3 = -1.51 eV.
  • E4=0.85E_4 = -0.85 eV.
  • E=0E_\infty = 0 (ionised).

The negative sign indicates binding (free electron has zero energy by convention).

Photon emission and absorption

When an electron transitions from level nin_i to nfn_f with ni>nfn_i > n_f, the atom emits a photon of energy:

Ephoton=EiEfE_{\text{photon}} = E_i - E_f

The photon frequency: f=Ephoton/hf = E_{\text{photon}} / h. The wavelength: λ=hc/Ephoton\lambda = hc / E_{\text{photon}}. Useful shortcut: hc1240hc \approx 1240 eV nm.

When an atom absorbs a photon whose energy exactly matches a transition EfEiE_f \to E_i, the electron is excited to the higher level.

Spectral series

Transitions ending at the same lower level form a series:

  • Lyman series (nf=1n_f = 1). Ultraviolet.
  • Balmer series (nf=2n_f = 2). Visible. H-alpha (3 to 2) at 656 nm; H-beta (4 to 2) at 486 nm; H-gamma (5 to 2) at 434 nm.
  • Paschen series (nf=3n_f = 3). Infrared.

The Balmer series is the canonical hydrogen spectrum visible in any high-voltage hydrogen discharge tube.

Cross-link: see the Rydberg spectrum calculator.

Emission and absorption spectra

Emission spectra. Hot, low-density gas emits photons during electron transitions to lower levels. Bright lines on dark background, each line a specific wavelength.

Absorption spectra. Continuous light passes through cool gas; specific wavelengths are absorbed. Dark lines on bright continuous spectrum.

Both spectra have the same line wavelengths because the same transitions are involved. The pattern of lines is the fingerprint of the element. Helium was first identified in solar spectra (1868) before being found on Earth.

Ionisation

When an electron absorbs enough energy to escape the atom (n=n = \infty), the atom is ionised. The minimum energy for ionisation from level nn is En|E_n|. For hydrogen ground state: 13.6 eV.

Multi-electron atoms

The Bohr model strictly applies to hydrogen (one electron). Multi-electron atoms have many more allowed transitions and more complex spectra, but the principle is the same: quantised energy levels, transitions emit/absorb photons.

QCE Physics Unit 4 uses the Bohr-level picture for hydrogen and assumes more complex atoms by analogy. Full quantum mechanics requires the Schrodinger equation, outside the scope of QCE.

Examples in context

Example 1. A Cairns hospital sodium-vapour lamp emits the 589 nm589 \text{ nm} yellow-orange line. Photon energy is E=hc/λ=(6.63×1034)(3×108)/(589×109)=3.38×1019 J=2.11 eVE = hc/\lambda = (6.63 \times 10^{-34})(3 \times 10^8)/(589 \times 10^{-9}) = 3.38 \times 10^{-19} \text{ J} = 2.11 \text{ eV}. This corresponds to the 3p3s3p \rightarrow 3s atomic transition in sodium, the Bohr model in practice.

Example 2. A QCAA Unit 4 photoelectric experiment uses UV from a mercury lamp (254 nm254 \text{ nm}, hf=4.88 eVh f = 4.88 \text{ eV}) on a ϕ=2.30 eV\phi = 2.30 \text{ eV} (sodium) photocathode. Stopping voltage Vs=(hfϕ)/e=2.58 VV_s = (h f - \phi)/e = 2.58 \text{ V}. The dot-point relationship Ek,max=hfϕE_{k,\max} = h f - \phi is exactly this analysis.

Try this

Q1. State the photoelectric equation and define each term. [2 marks]

  • Cue. Ek,max=hfϕE_{k,\max} = h f - \phi; work function ϕ\phi minimum energy to liberate electron.

Q2. Calculate the energy of a 600 nm600 \text{ nm} photon in joules and eV. [3 marks]

  • Cue. E=hc/λ=3.32×1019 J=2.07 eVE = hc/\lambda = 3.32 \times 10^{-19} \text{ J} = 2.07 \text{ eV}.

Q3. A photoelectric cell uses sodium (ϕ=2.30 eV\phi = 2.30 \text{ eV}) illuminated at 254 nm254 \text{ nm}. (a) Calculate the photon energy and the stopping voltage. (b) Determine the threshold frequency for sodium. (c) Explain why classical wave theory cannot account for the threshold. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Evaluation]

  • Cue. (a) hf=4.88 eVh f = 4.88 \text{ eV}, Vs=2.58 VV_s = 2.58 \text{ V}; (b) f0=ϕ/h=5.56×1014 Hzf_0 = \phi/h = 5.56 \times 10^{14} \text{ Hz}; (c) classical predicts intensity-driven not frequency-driven onset.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 QCAA-style5 marksLight of frequency 7.5×10147.5 \times 10^{14} Hz is incident on a metal with work function 2.42.4 eV. (a) Calculate photon energy in eV (use h=4.14×1015h = 4.14 \times 10^{-15} eV s). (b) Calculate the maximum kinetic energy of an ejected electron. (c) Calculate the threshold frequency for this metal.
Show worked answer →
(a) Photon energy
E=hf=4.14×1015×7.5×1014=3.105E = hf = 4.14 \times 10^{-15} \times 7.5 \times 10^{14} = 3.105 eV 3.1\approx 3.1 eV.
(b) Maximum KE
Ek,max=hfϕ=3.12.4=0.7E_{k,\max} = hf - \phi = 3.1 - 2.4 = 0.7 eV.
(c) Threshold frequency
f0=ϕ/h=2.4/4.14×1015=5.8×1014f_0 = \phi / h = 2.4 / 4.14 \times 10^{-15} = 5.8 \times 10^{14} Hz.

Markers reward correct application of each formula and consistent units throughout.

2023 QCAA-style4 marksCalculate the wavelength of the photon emitted when a hydrogen electron drops from n=4n = 4 to n=2n = 2. Use En=13.6/n2E_n = -13.6/n^2 eV.
Show worked answer →

E4=13.6/16=0.85E_4 = -13.6/16 = -0.85 eV.

E2=13.6/4=3.40E_2 = -13.6/4 = -3.40 eV.

ΔE=E4E2=0.85(3.40)=2.55\Delta E = E_4 - E_2 = -0.85 - (-3.40) = 2.55 eV.

Photon energy: Ephoton=2.55E_{\text{photon}} = 2.55 eV.

Wavelength: λ=hc/E\lambda = hc/E. Using hc=1240hc = 1240 eV nm:

λ=1240/2.55486\lambda = 1240 / 2.55 \approx 486 nm.

This is the H-beta line of the Balmer series (visible, blue-green).

Markers reward the energy-level subtraction, the photon-energy identification, and the hc/Ehc/E conversion to wavelength.

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