Photoelectric Effect
Section: Quantum Physics | Syllabus: Cambridge A Level Physics (9702)
Photoelectric Effect
Photoelectric emission is the emission of electrons from the surface of a metal when electromagnetic radiation of sufficiently high frequency is incident on it.
Key observations:
- Emission is instantaneous.
- Emission only occurs if the frequency of incident radiation is above a certain threshold frequency (f_0).
- If f < f_0, no electrons are emitted, regardless of intensity.
- The maximum kinetic energy of emitted electrons depends on frequency, not intensity.
- The rate of emission (photocurrent) depends on intensity.
Einstein’s Photoelectric Equation
Einstein explained the effect by proposing light consists of photons. A single photon interacts with a single electron on the surface (one-to-one interaction).
Conservation of Energy:
Photon Energy = Work Function + Kinetic Energy of Electronhf = Φ + (1/2)mv_max^2
Only electrons at the very surface require just Φ energy to escape. Electrons deeper inside require more energy, so they emerge with less kinetic energy. Hence, the equation gives the maximum kinetic energy.
Interpretation of the Equation
Rearranging for E_k(max):
E_k(max) = hf - ΦThis is in the form of a straight line equation y = mx + c, where:
- y-axis: Maximum Kinetic Energy (E_k(max))
- x-axis: Frequency (f)
- Gradient (m): Planck constant (h)
- y-intercept (c): -Φ (negative work function)
- x-intercept: Threshold frequency (f_0)
A straight line graph starting from a negative y-intercept (-Φ), crossing the x-axis at f0, with a positive gradient h.
Stopping Potential (V_s)
The maximum kinetic energy of the photoelectrons can be measured experimentally using a stopping potential.
Stopping Potential (V_s) The minimum negative (retarding) potential applied to the anode/collecting plate that prevents even the most energetic photoelectrons from reaching it.At the stopping potential, the work done by the electric field equals the maximum kinetic energy of the electrons:
E_k(max) = eV_sSubstituting into Einstein's equation:
eV_s = hf - Φ
This provides a method to determine Planck's constant. If we plot V_s against f:
- The equation is V_s = (h/e)f - (Φ/e)
- Gradient: (h/e)
- x-intercept: Threshold frequency (f_0)
- y-intercept: -(Φ/e)
Intensity vs Frequency
| Change | Effect on Max KE | Effect on Photocurrent (Rate of Emission) |
|---|---|---|
| Increase Frequency (at constant intensity) | Increases (each photon has more energy, leaving more for KE) | Decreases (higher f means fewer photons for same total power, so fewer electrons emitted) Careful: Assuming fixed Intensity W/m2 |
| Increase Intensity (at constant frequency > f_0) | No Change (photon energy unchanged) | Increases (more photons per second arrive, so more electrons emitted) |
Failures of Wave Theory
The classical wave theory of light could not explain the photoelectric effect. Here are the three key failures:
1. Existence of Threshold FrequencyWave Theory Prediction: Electrons should be emitted at any frequency, provided the intensity is high enough to supply the energy.
Observation: No emission occurs below a specific threshold frequency f_0, regardless of intensity.
2. Instantaneous EmissionWave Theory Prediction: At low intensities, there should be a time delay while electrons absorb enough energy from the wave wavefront to escape.
Observation: Emission is instantaneous (no time delay), even at very low intensities. (This implies one-to-one particle interaction).
3. Max KE depends on Frequency, not IntensityWave Theory Prediction: Increasing intensity (amplitude) should increase the kinetic energy of the emitted electrons.
Observation: Increasing intensity only increases the number of electrons (photocurrent), not their kinetic energy. Kinetic energy depends only on frequency.
Common TrapRemember that increasing intensity means more photons, NOT more energetic photons. Increasing frequency means more energetic photons.
Worked Examples
Worked Example: Maximum Kinetic Energy of PhotoelectronsQuestion: Calculate the maximum kinetic energy of electrons emitted when electromagnetic radiation of frequency 6.7 × 10^14 Hz is incident on a metal with a work function of 2.3 eV.
Solution1. Calculate photon energy: E = hf = (6.63 × 10^-34)(6.7 × 10^14) = 4.44 × 10^-19 J
2. Convert to eV: E = 4.44 × 10^-191.60 × 10^-19 = 2.78 eV
3. Apply Einstein's photoelectric equation: E_k(max) = hf - Φ
E_k(max) = 2.78 - 2.3 = 0.48 eV
Worked Example: Stopping PotentialQuestion: The stopping potential of electrons emitted from a metal is found to be 12.8 V. Calculate:
(a) the maximum kinetic energy of the photoelectrons in joules
(b) the maximum speed of the electrons. (m_e = 9.11 × 10^-31 kg)
(a) Using E_k(max) = eV_s:
E_k(max) = (1.60 × 10^-19)(12.8) = 2.05 × 10^-18 J
(b) Using E_k = (1/2)mv^2, rearrange for v:
v = √(2E_k/m) = √2 × 2.05 × 10^-189.11 × 10^-31
v = √4.50 × 10^12 = 2.12 × 10^6 m s^-1
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