| Syllabus Item | Content Covered in These Notes |
|---|---|
| 22.1 – Photon energy & momentum (recall & use equations) | Sections 1–3, derivation of \(p = h/\lambda\), constants table. |
| 22.2 – Photo‑electric effect (experiment, threshold, stopping potential) | Section 4, worked example, experimental note. |
| 22.3 – Wave‑particle duality & de Broglie wavelength | Section 5, de Broglie hypothesis and electron‑diffraction example. |
| 22.4 – Atomic energy‑level transitions & line spectra (apply \(hf = \Delta E\)) | Section 6, Balmer‑series derivation and example. |
| Quantity | Relation | Notes |
|---|---|---|
| Energy | \(E = h\nu\) | Planck’s quantisation; \(\nu\) = frequency (Hz) |
| Energy (wavelength form) | \(E = \dfrac{hc}{\lambda}\) | Using \(c = \lambda\nu\) |
| Momentum | \(p = \dfrac{h}{\lambda} = \dfrac{E}{c}\) | From relativistic energy–momentum relation for \(m_{0}=0\) |
| Frequency–wavelength link | \(c = \lambda\nu\) | Speed of light in vacuum |
\[
E^{2} = (pc)^{2} + (m_{0}c^{2})^{2}.
\]
With \(m_{0}=0\) this reduces to \(E = pc\).
| Symbol | Quantity | Value |
|---|---|---|
| \(h\) | Planck constant | \(6.626\times10^{-34}\ \text{J·s}\) |
| \(c\) | Speed of light in vacuum | \(2.998\times10^{8}\ \text{m·s}^{-1}\) |
| \(\lambda\) | Wavelength of a photon | Variable (m) |
| \(\nu\) | Frequency of a photon | Variable (Hz) |
| \(e\) | Elementary charge | \(1.602\times10^{-19}\ \text{C}\) |
\[
hf = \phi + \tfrac12 m{e}v{\max}^{2},
\]
where \(\phi\) is the work function.
\[
hf = \phi + eV_{s}.
\]
Problem: Light of wavelength \(250\ \text{nm}\) falls on a sodium surface (\(\phi = 2.28\ \text{eV}\)). Find the stopping potential.
\[
V_{s}= \frac{hf-\phi}{e}= \frac{4.97\ \text{eV}-2.28\ \text{eV}}{1\ \text{eV/e}} = 2.69\ \text{V}.
\]
\[
\lambda = \frac{h}{p}.
\]
Electrons accelerated through a potential \(V = 150\ \text{V}\) have kinetic energy \(eV\). Their momentum is
\[
p = \sqrt{2m_{e}eV},
\]
so the de Broglie wavelength is
\[
\lambda = \frac{h}{\sqrt{2m_{e}eV}} \approx 0.10\ \text{nm},
\]
comparable with inter‑atomic spacings, allowing diffraction.
\[
hf = E{i} - E{f}.
\]
\[
\frac{1}{\lambda}=R{Z}\!\left(\frac{1}{n{f}^{2}}-\frac{1}{n_{i}^{2}}\right),
\]
where \(R{Z}=R{\!H}Z^{2}\) and \(R_{\!H}=1.097\times10^{7}\ \text{m}^{-1}\).
\[
\frac{1}{\lambda}=R_{\!H}\!\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right),\qquad n=3,4,5,\dots
\]
\[
\frac{1}{\lambda}=1.097\times10^{7}\!\left(\frac{1}{4}-\frac{1}{9}\right)
=1.097\times10^{7}\times\frac{5}{36}=1.52\times10^{6}\ \text{m}^{-1},
\]
\[
\lambda = \frac{1}{1.52\times10^{6}} = 6.58\times10^{-7}\ \text{m}=658\ \text{nm}.
\]
\[
\Delta\lambda = \frac{h}{m_{e}c}\,(1-\cos\theta),
\]
demonstrating transfer of photon momentum to the electron.
A typical laboratory set‑up:
Combined standard uncertainty is typically \(\approx 3\%\).
A 650 nm laser delivers an average power of \(5.0\ \text{mW}\). Find the number of photons emitted per second.
\[
E = \frac{hc}{\lambda}= \frac{(6.626\times10^{-34})(2.998\times10^{8})}{6.50\times10^{-7}}
= 3.05\times10^{-19}\ \text{J}.
\]
\[
N = \frac{P}{E}= \frac{5.0\times10^{-3}}{3.05\times10^{-19}}
\approx 1.64\times10^{16}\ \text{s}^{-1}.
\]

A photon is a mass‑less quantum of electromagnetic radiation. Its energy and momentum are directly linked to its frequency and wavelength:
\[
E = h\nu = \frac{hc}{\lambda},\qquad
p = \frac{h}{\lambda} = \frac{E}{c}.
\]
These simple relations bridge the wave description (\(\lambda,\nu\)) and the particle description (energy, momentum) and underpin the photo‑electric effect, radiation pressure, Compton scattering, atomic line spectra and the broader concept of wave‑particle duality.
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