Free Pearson Edexcel International A Level Physics lesson

Photons, photoelectrons and atomic spectra

Use discrete energy transfers to explain observations a wave-only model misses.

What you’ll learn

  • Connect photon energy to frequency and wavelength.
  • Interpret threshold frequency, work function and photoelectron energy.
  • Calculate atomic transition frequencies and explain line spectra.

Two models answer different observations

Interference, diffraction and polarisation provide evidence for a wave description of electromagnetic radiation. Photoelectric observations require energy transfer in discrete packets called photons. These models developed because different experiments constrained what a successful explanation must predict; a new model did not make the earlier diffraction evidence disappear. A useful answer names the observation and explains why the proposed model accounts for it.

A photon has energy E = hf. Since electromagnetic waves in vacuum obey c = fλ, photon energy can also be written hc/λ. Shorter vacuum wavelength therefore means higher photon energy. The electronvolt is convenient for small energies: one electronvolt equals the work done moving a charge of magnitude e through one volt. Convert eV to joules before mixing with h expressed in joule seconds.

Beam power is total energy delivered per second. For monochromatic light, photon arrival rate is P/(hf), provided the stated power is the power carried by those photons. A brighter beam at fixed frequency contains more energy packets per second without increasing the energy of each packet. Detection and absorption can have efficiencies below one, so incident photon rate is not automatically equal to recorded electron count or measured electrical current.

Ephoton = hf = hc/λ; 1 eV = 1.60 × 10⁻¹⁹ J; photon rate = P/(hf)

Symbols and units

  • E is energy per photon in J; h is Planck’s constant in J s; f is frequency in Hz; λ is vacuum wavelength in m.
  • P is beam power in W and c is vacuum light speed in m s⁻¹.

How the equation explains the idea

  1. Combine E = hf with c = fλ to obtain E = hc/λ. Shorter wavelength means greater energy per photon.
  2. Power is energy per second, so photon arrival rate = P/E. At fixed power, doubling photon frequency doubles each photon’s energy and halves the number arriving per second.

Worked example

A monochromatic light pulse

  1. A photon has frequency 5.00 × 10¹⁴ Hz.
  2. E = 6.63 × 10⁻³⁴ × 5.00 × 10¹⁴ = 3.315 × 10⁻¹⁹ J.
  3. Dividing by 1.60 × 10⁻¹⁹ J per eV gives 2.07 eV.

The photon energy is set by frequency, independently of how many identical photons arrive.

Keep this idea. Choose wave or photon reasoning from the observation, and separate energy per photon from beam power.

One photon must overcome the work function

In the photoelectric effect, a surface absorbs electromagnetic radiation and emits electrons if individual photons transfer enough energy. The minimum surface-dependent energy required for escape is the work function φ. The threshold frequency f₀ satisfies hf₀ = φ. Below that frequency, increasing ordinary light intensity does not make each photon more energetic, so the simple one-photon process does not begin merely because more sub-threshold photons arrive.

For a photon that liberates an electron, energy conservation gives hf = φ + Kmax, where Kmax is the maximum emitted electron kinetic energy. Some electrons lose energy before escaping, so the emitted electrons need not all have that maximum. Increasing frequency above threshold raises Kmax. Increasing intensity at unchanged above-threshold frequency mainly raises the number of available photoelectrons per second, subject to collection and saturation conditions.

The observations of a threshold, essentially immediate emission under suitable above-threshold illumination, and maximum kinetic energy depending on frequency conflict with a simple classical wave picture in which an electron steadily accumulates arbitrarily small amounts of energy from any frequency. A Kmax versus f graph has slope h and frequency-axis intercept f₀. If stopping potential is measured, eVs equals Kmax, provided contact-potential and apparatus effects are appropriately considered.

hf = φ + Kmax; f₀ = φ/h; Kmax = eVs for the stopping-potential model

Symbols and units

  • φ is work function in J, h is Planck’s constant in J s, f is incident frequency in Hz and Kmax is maximum electron kinetic energy in J.
  • e is positive elementary charge in C; Vs is stopping-potential magnitude in V.

How the equation explains the idea

  1. A photon transfers hf. After paying the minimum escape energy φ, the maximum remaining kinetic energy is Kmax = hf − φ, only for f ≥ φ/h.
  2. The threshold is f₀ = φ/h. Increasing intensity below threshold does not make this single-photon model emit electrons. Above threshold the Kmax-versus-f slope is h.
  3. A retarding potential removes energy eVs, so at stopping eVs = Kmax.
The threshold is where emission can begin

Conditions. Single-photon photoelectric model with threshold f₀ = 5.00 × 10¹⁴ Hz, h = 6.626 × 10⁻³⁴ J s and e = 1.602 × 10⁻¹⁹ C. Only frequencies at or above threshold are plotted; below threshold there are no photoelectrons.

Frequency / 10¹⁴ Hz; Maximum kinetic energy / eV.

Kmax = h(f − f₀). The horizontal scale is in units of 10¹⁴ Hz and the vertical scale is eV, so the plotted slope is about 0.414 eV per 10¹⁴ Hz, not the numerical SI value of h. At 10 × 10¹⁴ Hz the energy is about 2.07 eV.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Energy after escape

  1. A photon supplies 4.0 eV to a surface of work function 2.5 eV.
  2. Maximum electron kinetic energy = 4.0 − 2.5 = 1.5 eV.
  3. In joules this is 1.5 × 1.60 × 10⁻¹⁹ = 2.4 × 10⁻¹⁹ J.

The work function is paid once in the energy balance for each emitted electron.

Keep this idea. Use energy per photon to decide emission, then use the surplus energy to find the maximum kinetic energy.

Read the gaps between atomic energy levels

An isolated atom has discrete allowed electron energies rather than a continuous choice of bound-state energies. When an electron moves from a higher to a lower allowed level, the atom can emit a photon with energy equal to the difference. Absorption is the reverse process when an incoming photon matches an allowed upward transition and the atom is in an appropriate initial state. The energy of a level alone is not the photon energy; the gap matters.

An emission spectrum therefore contains lines at particular frequencies rather than all frequencies. Different atoms have different level structures, so their line patterns help identify their presence in a source. A diagram often places bound levels at negative energy relative to the ionisation limit at zero. A transition from −2 eV to −5 eV releases 3 eV; the negative level labels do not mean that the emitted photon has negative energy.

Several downward routes may be possible from an excited state. A single electron can decay through intermediate levels, producing several photons whose total energy equals the overall energy drop, or take an allowed direct route with one larger-energy photon. Not every mathematically possible pair is necessarily an allowed transition in real atomic physics, so follow the allowed transitions stated in the question. Use ΔE = hf and convert eV consistently before finding frequency or wavelength.

Emitted photon energy = Ehigher − Elower = hf; absorption requires a matching allowed positive energy gap.

Symbols and units

  • Ehigher and Elower are allowed electron energies in the same energy unit; h is Planck’s constant in J s.
  • f is emitted photon frequency in Hz; use joules when dividing by h in SI units.

How the equation explains the idea

  1. An emission transfers the positive difference Ehigher − Elower into one photon. Negative level energies still give a positive difference for a downward transition.
  2. A transition from −2 eV to −5 eV emits 3 eV = 4.80 × 10⁻¹⁹ J using the rounded charge conversion; divide this difference by h, not either level separately.

Worked example

A two-step cascade

  1. An atom falls from −1.0 eV to −3.0 eV and then to −6.0 eV.
  2. The photons carry 2.0 eV and 3.0 eV respectively.
  3. Their total 5.0 eV equals the overall drop from −1.0 eV to −6.0 eV.

A cascade can share the total energy among photons without violating conservation.

Keep this idea. Subtract level energies in the correct order, then convert the positive gap into frequency or wavelength.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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