understand that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation
Energy and Momentum of a Photon – Cambridge A‑Level Physics 9702
Learning Objective
Understand that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation, and use the photon description of light to explain the observed phenomena.
1. Photons – Quantised Packets of Light
1.1 Energy of a Photon
A photon is a quantum of electromagnetic radiation. Its energy is directly proportional to its frequency (or inversely proportional to its wavelength):
\[
E = h\nu = \frac{hc}{\lambda}
\]
- Planck’s constant \(h = 6.626\times10^{-34}\ \text{J·s}\)
- Speed of light \(c = 3.00\times10^{8}\ \text{m·s}^{-1}\)
- \(\nu\) – frequency (Hz); \(\lambda\) – wavelength (m)
1.2 Momentum of a Photon
Because photons are mass‑less, their momentum follows from the relativistic relation \(E = pc\):
\[
p = \frac{E}{c} = \frac{h}{\lambda}
\]
Example: a photon of wavelength 500 nm carries
\[
p = \frac{6.626\times10^{-34}}{5.00\times10^{-7}}
= 1.33\times10^{-27}\ \text{kg·m·s}^{-1}.
\]
This tiny momentum underlies radiation‑pressure phenomena. For instance, sunlight exerts a pressure of about \(9.1\times10^{-6}\ \text{N·m}^{-2}\) on a perfectly absorbing surface, a principle exploited in solar‑sail spacecraft.
1.3 Wave‑Particle Duality (Syllabus 22.3)
The photon model provides a particle description of light, while diffraction and interference require a wave description. The photoelectric effect is classic evidence for the particle nature because:
- Emission occurs only when the photon energy exceeds a material‑specific work function.
- The kinetic energy of emitted electrons depends on the photon frequency, not on the light intensity.
At the same time, the same light produces interference patterns in double‑slit experiments, demonstrating its wave character. Together these observations illustrate wave‑particle duality, as required by the syllabus.
2. The Photoelectric Effect
2.1 Key Experimental Observations (Syllabus 22.2)
- Threshold frequency \(\nu0\): electrons are emitted only if \(\nu > \nu0\) (or \(\lambda < \lambda_0\)).
- Kinetic energy of the most energetic electrons increases linearly with \(\nu\) and is independent of light intensity.
- Increasing the intensity (at fixed \(\nu > \nu_0\)) raises the number of emitted electrons but does not change their maximum kinetic energy.
- Instantaneous emission: the delay between photon arrival and electron ejection is less than \(10^{-9}\ \text{s}\); for practical purposes the process is instantaneous.
2.2 Einstein’s Photo‑electric Equation
Einstein treated each incident photon as delivering its whole energy to a single electron:
\[
h\nu = \phi + K_{\max}
\]
- \(\phi\) – work function (minimum energy required to free an electron from the metal surface).
- \(K_{\max}\) – maximum kinetic energy of the emitted electron.
The work function is related to the threshold frequency by
\[
\phi = h\nu_0.
\]
2.3 Threshold Frequency and Threshold Wavelength
| Quantity | Expression | Comment |
|---|---|---|
| Threshold frequency \(\nu_0\) | \(\displaystyle \nu_0 = \frac{\phi}{h}\) | Minimum frequency required for emission. |
| Threshold wavelength \(\lambda_0\) | \(\displaystyle \lambda_0 = \frac{hc}{\phi}\) | Photons with \(\lambda > \lambda_0\) cannot eject electrons. |
2.4 Stopping Potential
In the standard apparatus a reverse (retarding) voltage \(V_s\) is applied between the photocathode and anode. The most energetic electrons are just prevented from reaching the anode when
\[
eVs = K{\max} = h\nu - \phi,
\]
where \(e = 1.602\times10^{-19}\ \text{C}\) is the elementary charge. Measuring \(V_s\) therefore gives a direct value for the kinetic energy of the fastest electrons.
3. Conditions for Photoelectron Emission
| Parameter | Requirement for Emission |
|---|---|
| Frequency \(\nu\) | \(\nu > \nu_0 = \dfrac{\phi}{h}\) |
| Photon energy \(E = h\nu\) | \(E > \phi\) (excess energy becomes kinetic energy) |
| Intensity | Controls the number of photons per second → controls the number of emitted electrons, not \(K_{\max}\). |
4. Momentum Transfer in Photoemission
When a photon ejects an electron, conservation of momentum must be satisfied:
\[
\underbrace{\frac{h}{\lambda}}{\text{photon momentum}} = \underbrace{pe}{\text{electron}} + \underbrace{p{\text{recoil}}}_{\text{metal lattice}}.
\]
Because the metal’s mass is enormously larger than that of the electron, the recoil momentum of the lattice is negligible for A‑Level calculations, but the principle reinforces the particle picture of light.
5. Example Calculations
Example 1 – Maximum Kinetic Energy
Find \(K_{\max}\) for a sodium surface (\(\phi = 2.28\ \text{eV}\)) illuminated with light of wavelength \(400\ \text{nm}\).
- Convert work function: \(\phi = 2.28\ \text{eV} = 2.28 \times 1.602\times10^{-19}\ \text{J} = 3.65\times10^{-19}\ \text{J}\).
- Photon energy:
\[
E = \frac{hc}{\lambda}
= \frac{6.626\times10^{-34}\times3.00\times10^{8}}{4.00\times10^{-7}}
= 4.97\times10^{-19}\ \text{J}.
\]
- Maximum kinetic energy:
\[
K_{\max}=E-\phi = 4.97\times10^{-19} - 3.65\times10^{-19}
= 1.32\times10^{-19}\ \text{J}.
\]
- In electron‑volts: \(K_{\max}= \dfrac{1.32\times10^{-19}}{1.602\times10^{-19}} \approx 0.82\ \text{eV}\).
Example 2 – Stopping Potential
Light of wavelength \(250\ \text{nm}\) shines on a metal with \(\phi = 3.0\ \text{eV}\). Determine the stopping potential required to just halt the most energetic electrons.
- Photon energy:
\[
E = \frac{hc}{\lambda}
= \frac{6.626\times10^{-34}\times3.00\times10^{8}}{2.50\times10^{-7}}
= 7.95\times10^{-19}\ \text{J}
= 4.96\ \text{eV}.
\]
- Maximum kinetic energy: \(K_{\max}=E-\phi = 4.96\ \text{eV} - 3.0\ \text{eV}=1.96\ \text{eV}\).
- Stopping potential: \(eVs = K{\max}\) → \(V_s = 1.96\ \text{V}\).
6. Typical Experimental Setup
7. Summary (Syllabus Checklist)
- Photon energy: \(E = h\nu = hc/\lambda\).
- Photon momentum: \(p = h/\lambda = E/c\).
- Work function \(\phi\) and threshold frequency \(\nu0\) (or wavelength \(\lambda0\)).
- Einstein’s equation \(h\nu = \phi + K_{\max}\) explains all observed features of the photoelectric effect.
- Stopping potential gives a direct measurement of \(K_{\max}\) and allows \(\phi\) to be determined experimentally.
- Intensity influences the number of emitted electrons, not their kinetic energy.
- Emission is essentially instantaneous (< 10⁻⁹ s), confirming the photon‑electron interaction is a single‑event process.
- Momentum conservation in photon–electron interactions reinforces the particle description of light.
- Combined with wave phenomena (diffraction, interference) the photoelectric effect illustrates wave‑particle duality (Syllabus 22.3).
8. Quick Revision Questions
- Calculate the threshold frequency and threshold wavelength for a metal with work function \(\phi = 4.5\ \text{eV}\).
- Light of wavelength \(300\ \text{nm}\) shines on a metal whose work function is \(2.5\ \text{eV}\). Determine:
- the maximum kinetic energy of the emitted electrons (in eV);
- the stopping potential required to stop them.
- Explain why increasing the intensity of light with \(\lambda = 600\ \text{nm}\) does not cause electron emission from a metal whose \(\phi = 2.0\ \text{eV}\).
- State two pieces of experimental evidence that support the particle nature of light and two that support its wave nature.