\(E = h\nu\), \(E = \dfrac{hc}{\lambda}\), \(p = \dfrac{E}{c}\), \(p = \dfrac{h}{\lambda}\).
The photon is introduced in Topic 7.4 (Electromagnetic spectrum). Its energy in eV links directly to:
\(1\;\text{eV}=1.602\times10^{-19}\;\text{J}\).
| Quantity | Formula | Units (SI) |
|---|---|---|
| Energy – frequency | \(E = h\nu\) | J (or eV) |
| Energy – wavelength | \(E = \dfrac{hc}{\lambda}\) | J (or eV) |
| Momentum – energy | \(E = pc\) | J·s m⁻¹ (or eV c⁻¹) |
| Momentum – wavelength | \(p = \dfrac{h}{\lambda}\) | kg·m s⁻¹ |
Special relativity gives the general energy–momentum relation
\[
E^{2}=p^{2}c^{2}+m^{2}c^{4}.
\]
For a photon the rest mass \(m=0\); therefore
\[
E^{2}=p^{2}c^{2}\;\Longrightarrow\;E = pc \quad (\text{positive root}).
\]
| Constant | Symbol | Value | Units |
|---|---|---|---|
| Speed of light | c | 2.998 × 10⁸ | m s⁻¹ |
| Planck constant | h | 6.626 × 10⁻³⁴ | J s |
| Reduced Planck constant | ħ | 1.055 × 10⁻³⁴ | J s |
| 1 electronvolt | 1 eV | 1.602 × 10⁻¹⁹ | J |
Convert joules to electronvolts by dividing by the conversion factor:
\[
E\;(\text{eV}) = \frac{E\;(\text{J})}{1.602\times10^{-19}}.
\]
Combining with \(E = hc/\lambda\) gives a handy formula when \(\lambda\) is in nanometres:
\[
\boxed{E\;(\text{eV}) = \frac{1240}{\lambda\;(\text{nm})}}.
\]
When frequency is given in terahertz (THz):
\[
\boxed{E\;(\text{eV}) = 4.1357\times10^{-3}\,\nu\;(\text{THz})}.
\]
From \(p = E/c\) the momentum can be expressed directly in terms of the photon energy:
\[
p = \frac{E}{c}.
\]
If the energy is given in eV it is often convenient to keep the speed of light in the denominator, defining the unit eV c⁻¹:
\[
p\;(\text{eV}\,c^{-1}) = \frac{E\;(\text{eV})}{c}\times\frac{1.602\times10^{-19}\;\text{J}}{1\;\text{eV}}.
\]
Numerically,
\[
p\;(\text{eV}\,c^{-1}) \approx 5.344\times10^{-28}\,E\;(\text{eV})\;\text{kg·m s}^{-1}.
\]
\[
\Delta\lambda = \frac{h}{m_ec}(1-\cos\theta),
\]
derived from conservation of energy and momentum for photons, confirming that photons carry momentum.
\[
eV_s = h\nu - \phi
\]
between stopping potential \(V_s\) and frequency \(\nu\) demonstrates quantised photon energy \(E = h\nu\) and the role of the work function \(\phi\) (both expressed in eV).
Problem: Find the energy (eV) and momentum (kg·m s⁻¹) of a photon with wavelength \(\lambda = 500\;\text{nm}\) (green light).
\[
E = \frac{hc}{\lambda}
= \frac{(6.626\times10^{-34})(2.998\times10^{8})}{500\times10^{-9}}
= 3.97\times10^{-19}\;\text{J}.
\]
\[
E\;(\text{eV}) = \frac{3.97\times10^{-19}}{1.602\times10^{-19}}
\approx 2.48\;\text{eV}.
\]
\[
p = \frac{E}{c}
= \frac{3.97\times10^{-19}}{2.998\times10^{8}}
= 1.33\times10^{-27}\;\text{kg·m s}^{-1}.
\]
\[
p\;(\text{eV}\,c^{-1}) = 5.34\times10^{-28}\times2.48
\approx 1.33\times10^{-27}\;\text{kg·m s}^{-1}.
\]
Problem: Light of wavelength 250 nm shines on a metal with work function \(\phi = 2.0\;\text{eV}\). Calculate the maximum kinetic energy of the emitted electrons and the required stopping potential.
\[
E = \frac{1240}{250}=4.96\;\text{eV}.
\]
\[
K_{\max}=E-\phi=4.96-2.0=2.96\;\text{eV}.
\]
\[
Vs = \frac{K{\max}}{e}=2.96\;\text{V}.
\]
| Quantity | Formula | SI Units | Common eV Form |
|---|---|---|---|
| Energy | \(E = h\nu = \dfrac{hc}{\lambda}\) | J or eV | \(E\;(\text{eV}) = \dfrac{1240}{\lambda\;(\text{nm})}\) |
| Momentum | \(p = \dfrac{E}{c} = \dfrac{h}{\lambda}\) | kg·m s⁻¹ | \(p\;(\text{eV}\,c^{-1}) = 5.34\times10^{-28}\,E\;(\text{eV})\) |
| Frequency | \(\nu = \dfrac{c}{\lambda}\) | Hz | \(E\;(\text{eV}) = 4.1357\times10^{-3}\,\nu\;(\text{THz})\) |
| Maximum kinetic energy (photo‑electric) | \(K_{\max}=E-\phi\) | J or eV | Use \(E\) in eV and \(\phi\) in eV. |
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