Discover how light quanta interact with electrons in metals
AQA A-Level Physics 2
📦Define the photon and calculate photon energy E = hf
🚪Define work function and threshold frequency
⚡Apply Einstein's photoelectric equation
🛑Explain stopping potential and its significance
🔬Explain observations of the photoelectric effect
📏Convert between eV and joules for photon energies
Photons and Photon Energy
Light and all electromagnetic radiation consists of discrete packets of energy called photons. This was proposed by Einstein in 1905. A photon is a quantum of electromagnetic radiation.
E = hf = hc/λ
h = 6.63 × 10⁻³⁴ J s (Planck's constant)
f = frequency (Hz); λ = wavelength (m); c = 3.00 × 10⁸ m s⁻¹
Electronvolt (eV): The energy gained by an electron accelerated through a potential difference of 1 volt. 1 eV = 1.60 × 10⁻¹⁹ J. Useful for expressing energies of photons and particles at atomic scales.
Type of EM radiation
Typical frequency / Hz
Photon energy / eV
Radio
10⁶
4 × 10⁻⁹
Visible light
~5 × 10¹⁴
~2
UV
10¹⁵
~4
X-rays
10¹⁸
~4000
The Photoelectric Effect
When electromagnetic radiation of sufficient frequency shines on a metal surface, electrons are emitted. This is the photoelectric effect. Key observations:
Electrons are emitted only if the frequency of light exceeds a minimum value called the threshold frequency f₀ — regardless of intensity.
Below f₀, no electrons are emitted no matter how intense the light.
Above f₀, emission is instantaneous — no time delay.
Increasing the intensity (at constant frequency above f₀) increases the number of electrons emitted per second, but NOT their maximum kinetic energy.
Increasing the frequency increases the maximum kinetic energy of emitted electrons.
The classical wave model of light cannot explain these observations. It predicts that any frequency would eventually emit electrons given enough time/intensity. The photon model explains all observations correctly.
Work Function and Einstein's Equation
Work function φ: The minimum energy needed to remove an electron from the surface of a metal. It is a property of the metal.
Threshold frequency f₀: The minimum frequency of light that will cause photoelectric emission: φ = hf₀, so f₀ = φ/h.
The photon gives all its energy to one electron. Some energy φ is used to free the electron from the metal surface. Any remaining energy appears as kinetic energy of the emitted electron. The maximum KE is for electrons emitted from the surface with no energy lost within the metal.
This is a one-photon-one-electron interaction. The photon is absorbed entirely, not split between electrons. This is why intensity (more photons, same energy each) increases photocurrent but not maximum KE.
Stopping Potential
Stopping potential V_s: The minimum negative potential applied to the collector to stop all emitted electrons from reaching it, even the fastest ones.
eV_s = E_k(max) = hf − φ
So: V_s = (hf − φ) / e
Stopping potential is measured by applying a reverse voltage and finding the value at which photocurrent just reaches zero. Plotting V_s against f gives a straight line:
This graph gives an experimental method to determine Planck's constant h. The gradient of V_s vs f = h/e = 6.63 × 10⁻³⁴ / 1.60 × 10⁻¹⁹ = 4.14 × 10⁻¹⁵ V s.
Light of wavelength 250 nm shines on a sodium surface (work function φ = 2.28 eV). Calculate the maximum kinetic energy of emitted electrons in joules and eV.
f₀ = 1.04 × 10¹⁵ Hz (UV range — visible light cannot cause emission from zinc)
A stopping potential of 1.20 V is measured for light of frequency 6.80 × 10¹⁴ Hz. Calculate Planck's constant from this data.
1eV_s = hf − φ. We need another frequency to find h graphically, but with a known work function we can use a single measurement. Assume φ for caesium = 2.00 eV = 3.20 × 10⁻¹⁹ J.
h ≈ 7.5 × 10⁻³⁴ J s (differs from 6.63 × 10⁻³⁴ due to approximate φ used)
Explain, using the photon model, why doubling the intensity of light below the threshold frequency still produces no photoelectrons.
1Intensity is proportional to the number of photons per second per unit area — more photons, but each photon has the same energy E = hf.
2Below threshold: hf < φ, so no single photon has enough energy to free an electron from the surface.
3The interaction is one-photon-one-electron. Two low-energy photons cannot combine to release one electron.
More photons → more collisions, but each carries insufficient energy → no emission regardless of intensity
1. Which variable determines whether photoelectric emission occurs?
Photoelectric emission occurs only if the photon frequency exceeds the threshold frequency (f ≥ f₀). Intensity, duration and amplitude are irrelevant to whether emission occurs.
2. Light of frequency 8.0 × 10¹⁴ Hz has what photon energy? (h = 6.63 × 10⁻³⁴ J s)
4. If intensity of light above f₀ is doubled, what happens to the maximum kinetic energy of emitted electrons?
E_k(max) = hf − φ depends only on photon frequency and work function. Doubling intensity doubles the number of photons (and thus the current), but each photon still has the same energy hf → same maximum KE.
5. A stopping potential of 0.80 V is needed for light of wavelength 400 nm on a metal. Find the work function in eV. (hc = 1.99 × 10⁻²⁵ J m)
1. A student plots stopping potential V_s (y-axis) against frequency f (x-axis) for several different metals and frequencies. The graph is a straight line. The gradient is 4.0 × 10⁻¹⁵ V s and the x-intercept is 5.5 × 10¹⁴ Hz. Find: (a) Planck's constant h, (b) the work function φ in eV.
2. A photon of 5.0 eV strikes a metal with work function 3.5 eV. (a) Calculate the maximum speed of emitted electrons. (b) Calculate the de Broglie wavelength of these electrons. (m_e = 9.11 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s)