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Topic 2: Quantum theory

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

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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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=6.626Γ—10βˆ’34h = 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/Ξ»h/\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 Ο•\phi (the work function) to escape and keeps the remainder as kinetic energy:

Ek,max⁑=hfβˆ’Ο•E_{k,\max} = hf - \phi

  • Below threshold (hf<Ο•hf < \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=Ο•hf_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⁑=hfβˆ’Ο•e 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=nℏL = n \hbar where ℏ=h/(2Ο€)\hbar = h / (2\pi).

For hydrogen, the energy levels are:

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

for n=1,2,3,…n = 1, 2, 3, \ldots.

  • IMATH_24 eV (ground state).
  • IMATH_25 eV.
  • IMATH_26 eV.
  • IMATH_27 eV.
  • IMATH_28 (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=Eiβˆ’EfE_{\text{photon}} = E_i - E_f

The photon frequency: f=Ephoton/hf = E_{\text{photon}} / h. The wavelength: Ξ»=hc/Ephoton\lambda = hc / E_{\text{photon}}. Useful shortcut: hcβ‰ˆ1240hc \approx 1240 eV nm.

When an atom absorbs a photon whose energy exactly matches a transition Ef→EiE_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.

Common errors

Confusing photon energy with electron kinetic energy. Ephoton=hfE_{\text{photon}} = hf is fixed by the light. Ek,max⁑=hfβˆ’Ο•E_{k,\max} = hf - \phi depends on both the photon and the metal.

Forgetting the negative sign on bound energy levels. EnE_n is negative. The photon emitted has positive energy Eiβˆ’EfE_i - E_f, where both are negative and Ei>EfE_i > E_f algebraically (less negative).

Treating intensity as energy per photon. Intensity is photons per second per area. Energy per photon is fixed by frequency.

Applying hc=1240hc = 1240 eV nm without unit check. This works only when Ξ»\lambda is in nm and EE is in eV. Mixed units fail.

Confusing emission and absorption directions. Emission: higher to lower (energy out). Absorption: lower to higher (energy in).

In one sentence

Quantum theory in QCE Physics Unit 4 is built around the photon model (E=hfE = hf for each quantum of light), the photoelectric effect (electrons ejected above threshold with Ek,max⁑=hfβˆ’Ο•E_{k,\max} = hf - \phi), and the Bohr model of the atom (quantised energy levels with transitions producing or absorbing photons of energy Ξ”E=Eiβˆ’Ef\Delta E = E_i - E_f, generating the line spectra used to identify elements).

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2024 QCAA-style5 marksLight of frequency $7.5 \times 10^{14}$ Hz is incident on a metal with work function $2.4$ eV. (a) Calculate photon energy in eV (use $h = 4.14 \times 10^{-15}$ eV s). (b) Calculate the maximum kinetic energy of an ejected electron. (c) Calculate the threshold frequency for this metal.
Show worked answer β†’

(a) Photon energy. E=hf=4.14Γ—10βˆ’15Γ—7.5Γ—1014=3.105E = hf = 4.14 \times 10^{-15} \times 7.5 \times 10^{14} = 3.105 eV β‰ˆ3.1\approx 3.1 eV.

(b) Maximum KE. Ek,max⁑=hfβˆ’Ο•=3.1βˆ’2.4=0.7E_{k,\max} = hf - \phi = 3.1 - 2.4 = 0.7 eV.

(c) Threshold frequency. f0=Ο•/h=2.4/4.14Γ—10βˆ’15=5.8Γ—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 $n = 4$ to $n = 2$. Use $E_n = -13.6/n^2$ eV.
Show worked answer β†’

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

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

Ξ”E=E4βˆ’E2=βˆ’0.85βˆ’(βˆ’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: Ξ»=hc/E\lambda = hc/E. Using hc=1240hc = 1240 eV nm:

Ξ»=1240/2.55β‰ˆ486\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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