explain photoelectric emission in terms of photon energy and work function energy

Energy and Momentum of Photons – Photo‑electric Emission (Cambridge 22.1)

1. Photon Energy and Momentum

  • Energy: \(E = h\nu = \dfrac{hc}{\lambda}\)
  • Momentum: \(p = \dfrac{E}{c} = \dfrac{h}{\lambda}\)

Unit‑check – 1 eV = \(1.602\times10^{-19}\) J.

When using electron‑volts, the photon energy can be written simply as

\(E\;(\text{eV}) = \dfrac{1240}{\lambda\;(\text{nm})}\).

Photon momentum in SI units: \(p\;(\text{kg·m·s}^{-1}) = \dfrac{h}{\lambda}\).

SymbolQuantityValue (SI)
\(h\)Planck’s constant\(6.626\times10^{-34}\ \text{J·s}\)
\(c\)Speed of light in vacuum\(3.00\times10^{8}\ \text{m·s}^{-1}\)
\(\nu\)Frequency of the radiationHz
\(\lambda\)Wavelength of the radiationm

Work Function and Threshold Frequency (Cambridge 22.2)

  • The work function \(\phi\) is the minimum energy required to liberate an electron from the surface of a particular metal.
  • It is material‑specific and is usually expressed in electron‑volts (eV).
  • Relation to the threshold frequency \(\nu{0}\) (or wavelength \(\lambda{0}\)):

    \[

    \phi = h\nu{0} = \frac{hc}{\lambda{0}}

    \]

  • If the incident light has \(\nu < \nu{0}\) (or \(\lambda > \lambda{0}\)) no electrons are emitted, regardless of the light intensity.

Einstein Photo‑electric Equation (Cambridge 22.2)

Einstein’s photo‑electric equation (the syllabus code for this relation is 22.2):

\[

h\nu = \phi + K_{\max}

\]

In a stopping‑potential experiment the maximum kinetic energy is measured via the retarding voltage \(V_{0}\):

\[

eV{0}=K{\max}=h\nu-\phi

\]

  • \(e\) – elementary charge (\(1.602\times10^{-19}\) C)
  • \(V_{0}\) – stopping (retarding) potential (V)

Rearrangements for common exam tasks (AO2):

  • Threshold frequency: \(\displaystyle \nu_{0}= \frac{\phi}{h}\)
  • Threshold wavelength: \(\displaystyle \lambda_{0}= \frac{hc}{\phi}\)
  • Stopping potential from measured \(K{\max}\): \(\displaystyle V{0}= \frac{K_{\max}}{e}\)

Effect of Light Intensity (Cambridge 22.2)

  • The kinetic energy of the emitted electrons depends only on the frequency of the incident photons, not on the intensity.
  • Increasing the light intensity (i.e. the number of photons per unit time) increases the photocurrent proportionally, because more electrons are emitted, but \(K_{\max}\) remains unchanged.
  • Quantitatively: if the intensity is increased by a factor \(n\), the photocurrent \(I\) increases by the same factor \(n\) while \(K_{\max}\) stays the same.

Evidence for Wave‑Particle Duality (Cambridge 22.3)

  • Threshold frequency: Only photons with energy \(h\nu\ge\phi\) can cause emission – a particle‑like property.
  • Instantaneous emission: Electrons are ejected without any measurable time lag, contradicting a wave‑based energy‑accumulation model.
  • Independence of intensity: Changing intensity changes the number of emitted electrons but not their kinetic energy, again supporting a photon description.
  • These observations together confirm Einstein’s quantum (photon) model while the wave description still explains interference and diffraction phenomena.

Practical Implications and Applications

  • Predicting the longest wavelength that can cause emission for a given metal:

    \[

    \lambda_{\max}= \frac{hc}{\phi}

    \]

  • Photoelectron spectroscopy – measuring \(K{\max}\) (or \(V{0}\)) to determine binding energies of electrons in atoms, molecules or solids.
  • Solar‑cell design – choosing electrode materials with appropriate work functions to maximise conversion of photon energy into electrical current.
  • Photomultiplier tubes, night‑vision devices and image‑intensifier tubes rely on the photo‑electric effect.

Example Calculation

For sodium metal the work function is \(\phi = 2.28\ \text{eV}\).

  1. Threshold wavelength \(\lambda_{0}\):

    \[

    \lambda_{0}= \frac{hc}{\phi}= \frac{(6.626\times10^{-34})(3.00\times10^{8})}{2.28\times1.602\times10^{-19}}

    \approx 5.44\times10^{-7}\ \text{m}=544\ \text{nm}

    \]

  2. Photon energy for \(\lambda = 400\ \text{nm}\):

    \[

    E = \frac{hc}{\lambda}= \frac{(6.626\times10^{-34})(3.00\times10^{8})}{4.00\times10^{-7}}

    =4.97\ \text{eV}

    \]

  3. Maximum kinetic energy:

    \[

    K_{\max}=E-\phi = 4.97\ \text{eV} - 2.28\ \text{eV}=2.69\ \text{eV}

    \]

  4. Stopping potential:

    \[

    V{0}= \frac{K{\max}}{e}= \frac{2.69\ \text{eV}}{1\ \text{eV·V}^{-1}} = 2.69\ \text{V}

    \]

Summary Table

QuantitySymbolExpressionUnits
Photon energy\(E\)\(h\nu = \dfrac{hc}{\lambda}\)J (or eV)
Photon momentum\(p\)\(\dfrac{h}{\lambda}= \dfrac{E}{c}\)kg·m·s⁻¹ (or N·s)
Work function\(\phi\)\(h\nu{0}= \dfrac{hc}{\lambda{0}}\)J (or eV)
Maximum kinetic energy\(K_{\max}\)\(h\nu-\phi = eV_{0}\)J (or eV)
Stopping potential\(V_{0}\)\(\displaystyle V{0}= \frac{K{\max}}{e}\)V

Suggested Diagram

Energy diagram of the photo‑electric effect: a photon of energy \(h\nu\) strikes a metal surface, overcomes the work‑function barrier \(\phi\), and an electron is emitted with kinetic energy \(K{\max}\). The stopping potential \(V{0}\) is shown as the retarding voltage required to halt the most energetic electrons.