recall and use hf = Φ + 21mvmax2

Energy and Momentum of a Photon – Cambridge AS & A Level Physics (9702)

1. Photon – the quantum of electromagnetic radiation

• A photon is the indivisible quantum (or “particle”) of an electromagnetic wave.
• Each photon carries a fixed amount of energy that depends only on the wave’s frequency, and a corresponding amount of momentum.
• The particle picture explains the photoelectric effect and radiation pressure; the wave picture explains interference and diffraction.

2. Core formulae (Syllabus 22.1)

• Energy of a photon

  \[

  E = hf

  \]

  • \(h = 6.626\times10^{-34}\ \text{J·s}\) (Planck’s constant)
  • \(f\) – frequency (Hz)

• Momentum of a photon

  \[

  p = \frac{E}{c}= \frac{h}{\lambda}

  \]

  • \(c = 3.00\times10^{8}\ \text{m·s}^{-1}\) (speed of light in vacuum)
  • \(\lambda\) – wavelength (m)

2.1 Quick derivation

From the wave relation \(c = f\lambda\) we have \(f = c/\lambda\). Substituting into \(E = hf\) gives

\[

E = h\frac{c}{\lambda}\quad\Longrightarrow\quad p = \frac{E}{c}= \frac{h}{\lambda}.

\]

2.2 Relativistic link (A‑Level depth)

The photon momentum formula also follows from the relativistic energy‑momentum relation

\[

E^{2} = (pc)^{2} + (m_{0}c^{2})^{2}.

\]

For a photon \(m_{0}=0\), so \(E = pc\) and therefore \(p = E/c\).

2.3 Unit‑conversion reminder

• Energy is often given in electron‑volts (eV). Convert to joules before using \(p = E/c\):

  \[

  1\ \text{eV}=1.602\times10^{-19}\ \text{J}.

  \]

• Alternatively, use the combined factor

  \[

  p\ (\text{kg·m·s}^{-1}) = \frac{E\ (\text{eV})}{c}\times1.602\times10^{-19}.

  \]

3. Radiation pressure – a practical use of photon momentum (optional, Syllabus 22.1)

When a stream of photons strikes a surface, each photon transfers its momentum \(p\). If \(N\) photons strike the surface each second, the average force is

\[

F = p\,\dot N = \frac{E}{c}\,\dot N = \frac{I\,A}{c},

\]

where \(I\) is the light intensity (W m⁻²) and \(A\) the illuminated area (m²).

This principle underlies solar sails, laser‑driven spacecraft and optical tweezers.

4. Photoelectric effect – quantitative use of the photon equation (Syllabus 22.2)

\[

hf = \Phi + \frac{1}{2}mv_{\max}^{2}

\]

• \(\Phi\) – work function, the minimum energy required to liberate an electron from the metal surface.

  • Units: joules (J) or electron‑volts (eV). Use \(1\ \text{eV}=1.602\times10^{-19}\ \text{J}\) for conversion.
  • Only electrons that reach the surface can escape; the work function represents the energy needed to overcome the surface binding.

• \(m = 9.11\times10^{-31}\ \text{kg}\) – electron mass.
• \(v_{\max}\) – maximum speed of the emitted electrons.
• Threshold frequency: \(\displaystyle f{\text{th}} = \frac{\Phi}{h}\) (or threshold wavelength \(\displaystyle \lambda{\text{th}} = \frac{hc}{\Phi}\)).

4.1 Why intensity affects the number of electrons but not \(v_{\max}\)

• Each photon carries a single quantum of energy \(hf\). An electron can absorb at most one photon; it either receives the full energy \(hf\) or none.
• Increasing the intensity means more photons per second, so more electrons are emitted, but the energy per photon (and therefore the maximum kinetic energy) stays the same.

4.2 Kinetic‑energy‑vs‑frequency graph (Syllabus 22.2)

4.3 Worked example – finding \(v_{\max}\)

Metal work function \(\Phi = 2.2\ \text{eV}\). Light wavelength \(\lambda = 400\ \text{nm}\).

1. Photon energy:

   \[

   E = \frac{hc}{\lambda}= \frac{1240\ \text{eV·nm}}{400\ \text{nm}} = 3.10\ \text{eV}.

   \]

2. Insert into the photoelectric equation:

   \[

   3.10\ \text{eV}=2.20\ \text{eV}+ \frac{1}{2}mv_{\max}^{2}.

   \]

3. Maximum kinetic energy:

   \[

   K_{\max}=3.10-2.20=0.90\ \text{eV}=0.90\times1.602\times10^{-19}=1.44\times10^{-19}\ \text{J}.

   \]

4. Maximum speed:

   \[

   v{\max}= \sqrt{\frac{2K{\max}}{m}}

   =\sqrt{\frac{2(1.44\times10^{-19})}{9.11\times10^{-31}}}

   =5.6\times10^{5}\ \text{m·s}^{-1}.

   \]

5. Numerical example – photon energy and momentum (Syllabus 22.1)

Calculate the energy and momentum of a photon with wavelength \(\lambda = 500\ \text{nm}\) (green light).

1. \(\lambda = 5.00\times10^{-7}\ \text{m}\).
2. Frequency: \(\displaystyle f = \frac{c}{\lambda}= \frac{3.00\times10^{8}}{5.00\times10^{-7}} = 6.00\times10^{14}\ \text{Hz}\).
3. Energy: \(\displaystyle E = hf = (6.626\times10^{-34})(6.00\times10^{14}) = 3.98\times10^{-19}\ \text{J}=2.48\ \text{eV}.\)
4. Momentum (using \(p=E/c\) or \(p=h/\lambda\)):

   \[

   p = \frac{E}{c}= \frac{3.98\times10^{-19}}{3.00\times10^{8}} = 1.33\times10^{-27}\ \text{kg·m·s}^{-1}.

   \]

6. Photon‑energy table for common wavelengths

7. Wave‑particle duality (Syllabus 22.3)

• Particle evidence: Photoelectric effect – each electron absorbs a single photon of energy \(hf\); kinetic energy depends on frequency, not intensity.
• Wave evidence: Young’s double‑slit experiment, diffraction gratings, and electron diffraction – the observed patterns require a wavelength description.

7.1 de Broglie wavelength

Louis de Broglie extended the wave‑particle idea to matter particles:

\[

\lambda = \frac{h}{p},

\qquad p = mv\ (\text{non‑relativistic}).

\]

7.2 Quantitative example – electron accelerated through 150 V

1. Kinetic energy gained: \(K = eV = (1.602\times10^{-19})(150)=2.40\times10^{-17}\ \text{J}\).
2. Momentum:

   \[

   p = \sqrt{2mK}= \sqrt{2(9.11\times10^{-31})(2.40\times10^{-17})}=9.34\times10^{-24}\ \text{kg·m·s}^{-1}.

   \]

3. de Broglie wavelength:

   \[

   \lambda = \frac{h}{p}= \frac{6.626\times10^{-34}}{9.34\times10^{-24}} = 7.1\times10^{-11}\ \text{m}=0.071\ \text{nm}.

   \]

4. This wavelength is comparable to inter‑atomic spacings, explaining why electrons of a few‑hundred‑volt energy can produce diffraction patterns.

8. Summary of key points

• A photon is the indivisible quantum of an EM wave; its energy and momentum are

  \[

  E = hf,\qquad p = \frac{E}{c}= \frac{h}{\lambda}.

  \]

  The relation \(p = E/c\) follows directly from the relativistic energy‑momentum equation for a mass‑less particle.

• Radiation pressure arises from the transfer of photon momentum: \(F = I A /c\).
• Photoelectric equation: \(hf = \Phi + \tfrac12 mv_{\max}^{2}\).

  • Threshold frequency \(f{\text{th}} = \Phi/h\) (or \(\lambda{\text{th}} = hc/\Phi\)).
  • \(\Phi\) is expressed in joules or electron‑volts; convert using \(1\ \text{eV}=1.602\times10^{-19}\ \text{J}\).
  • Electrons must overcome the surface work function to escape.
  • Increasing light intensity raises the number of emitted electrons but does not change \(v_{\max}\) because each electron absorbs only one photon.

• Wave‑particle duality is supported by both particle evidence (photoelectric effect) and wave evidence (interference, diffraction).

  The de Broglie relation \(\lambda = h/p\) links momentum to wavelength for all matter, predicting observable diffraction for electrons, neutrons, etc.

9. Practice questions (with brief hints)

1. Calculate the momentum of a \(1.0\ \mu\text{m}\) infrared photon.

   Hint: Find \(E = hc/\lambda\), convert to joules, then use \(p = E/c\).

2. A metal has \(\Phi = 1.8\ \text{eV}\). Light of wavelength \(250\ \text{nm}\) shines on it. Determine the maximum kinetic energy of the emitted electrons in eV.

   Hint: Compute photon energy \(E = hc/\lambda\) in eV, then \(K_{\max}=E-\Phi\).

3. Explain why increasing the intensity of light (while keeping wavelength constant) increases the number of emitted electrons but does not affect their maximum speed \(v_{\max}\).

   Hint: Relate intensity to photon flux; each electron absorbs a single photon.

4. Sketch the kinetic‑energy‑vs‑frequency graph for the photoelectric effect, label the threshold frequency and the slope.

   Hint: Vertical axis in eV, horizontal axis in Hz; slope = \(h\).

5. Calculate the de Broglie wavelength of an electron accelerated through a potential difference of \(150\ \text{V}\).

   Hint: Use \(K = eV\), then \(p = \sqrt{2mK}\) and \(\lambda = h/p\).