The Quantum Model of Light

Planck's revolutionary idea was that energy is quantised: an oscillator can only emit or absorb whole packets $E = hf$. This discreteness suppresses high-frequency emission and resolved the ultraviolet catastrophe.

$E = hf = (6.63\times10^{-34})(5.0\times10^{14}) = 3.3\times10^{-19}\text{ J}$.

Classical theory predicted the emitted intensity would rise without limit as the wavelength got shorter (into the UV), which is impossible. Planck's quantisation $E = hf$ suppressed high-frequency emission and matched the observed curve.

Wien's law: $\lambda_{\max} = \dfrac{b}{T} = \dfrac{2.898\times10^{-3}}{7000} = 4.14\times10^{-7}\text{ m} = 414\text{ nm}$.

By Wien's law a shorter peak wavelength means a higher temperature. A hotter black body's curve lies above the cooler one at every wavelength, so it is both hotter and brighter overall.

Emission requires $hf \ge \phi$, i.e. a frequency above the threshold $f_0$. Below $f_0$ no electrons escape however bright the light; above it, even dim light ejects electrons immediately.

$E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times10^{-34})(3.0\times10^8)}{200\times10^{-9}} = \dfrac{1.989\times10^{-25}}{2.0\times10^{-7}} = 9.9\times10^{-19}\text{ J}$ ($\approx 6.2\text{ eV}$).

More intensity means more photons per second, so more electrons per second (a larger current). But each electron's $K_{\max} = hf - \phi$ depends only on frequency, so its maximum KE is unchanged.

The retarding voltage does work against the electrons. At the stopping voltage $V_s$ even the fastest electrons are turned back and the current falls to zero, so $K_{\max} = eV_s$.

$K_{\max} = hf - \phi = 5.0 - 3.0 = 2.0\text{ eV}$. The photon's energy is split between freeing the electron (the work function) and the electron's kinetic energy.

At threshold $hf_0 = \phi$, so $f_0 = \dfrac{\phi}{h} = \dfrac{2.5}{4.14\times10^{-15}} = 6.0\times10^{14}\text{ Hz}$.

$\phi = 2.5\text{ eV} = 4.0\times10^{-19}\text{ J}$. Then $\lambda_0 = \dfrac{hc}{\phi} = \dfrac{1.989\times10^{-25}}{4.0\times10^{-19}} = 4.97\times10^{-7}\text{ m} = 497\text{ nm}$ (green). Longer wavelengths cannot eject electrons.

$K_{\max} = eV_s$, so $V_s = \dfrac{K_{\max}}{e}$. With $K_{\max}$ expressed in eV the stopping voltage in volts is numerically equal: $V_s = 2.0\text{ V}$.

Photon energy $E = \dfrac{hc}{\lambda} = \dfrac{1.989\times10^{-25}}{250\times10^{-9}} = 7.96\times10^{-19}\text{ J} = 4.97\text{ eV}$. Then $K_{\max} = E - \phi = 4.97 - 3.0 \approx 2.0\text{ eV}$.

Wien's law gives $\lambda_{\max} \propto \dfrac{1}{T}$. Halving the peak wavelength doubles the temperature.

Emission needs the photon energy to exceed $\phi = 2.3\text{ eV}$. Red $1.9 < 2.3$ (no); green $2.4 > 2.3$ (yes); UV $4.1 > 2.3$ (yes). Intensity is irrelevant to whether emission occurs.

$K_{\max} = hf - \phi$. At double the frequency $K_{\max}' = 2hf - \phi$, which exceeds $2K_{\max} = 2hf - 2\phi$ by $\phi$. Because of the $-\phi$ offset, $K_{\max}$ is not proportional to $f$, so doubling $f$ more than doubles it.

Rearranged, $V_s = \dfrac{h}{e}f - \dfrac{\phi}{e}$. So the gradient is $\dfrac{h}{e}$ (the same for every metal), the x-intercept is the threshold frequency $f_0$, and the y-intercept is $-\dfrac{\phi}{e}$.

The gradient is $\dfrac{h}{e} = \dfrac{2.57 - 0.50}{1.0\times10^{15} - 5.0\times10^{14}} = \dfrac{2.07}{5.0\times10^{14}} = 4.14\times10^{-15}\text{ V/Hz}$. Then $h = \text{slope}\times e = (4.14\times10^{-15})(1.6\times10^{-19}) = 6.6\times10^{-34}\text{ J·s}$.

$eV_s = hf - \phi$. With $K_{\max} = eV_s = 1.6\text{ eV}$: $\phi = hf - K_{\max} = (4.14\times10^{-15})(1.0\times10^{15}) - 1.6 = 4.14 - 1.6 = 2.5\text{ eV}$.

Twice the intensity means twice as many photoelectrons per second, so the saturation current doubles. But $K_{\max}$, and hence $V_s$, depends only on the frequency, so the stopping voltage is unchanged (both curves cut the axis at the same $-V_s$).

Energy per photon $E = \dfrac{hc}{\lambda} = \dfrac{1.989\times10^{-25}}{660\times10^{-9}} = 3.0\times10^{-19}\text{ J}$. Photons per second $= \dfrac{P}{E} = \dfrac{3.0\times10^{-3}}{3.0\times10^{-19}} \approx 1.0\times10^{16}$.

The wave model predicts that a bright enough beam of any frequency should eventually free electrons (after a delay), with more KE for brighter light. Experiment shows a sharp threshold frequency, instantaneous emission and $K_{\max} = hf - \phi$, all explained by photons of energy $hf$. Reflection, interference, polarisation and refraction are the wave model's successes.

$K_{\max} = 2.0\text{ eV} = 3.2\times10^{-19}\text{ J}$. From $K_{\max} = \tfrac12 m v^2$: $v = \sqrt{\dfrac{2K_{\max}}{m_e}} = \sqrt{\dfrac{2(3.2\times10^{-19})}{9.1\times10^{-31}}} = \sqrt{7.0\times10^{11}} \approx 8.4\times10^{5}\text{ m/s}$.

Photon energy at $300\text{ nm}$: $E = \dfrac{hc}{\lambda} = \dfrac{1.989\times10^{-25}}{300\times10^{-9}} = 6.63\times10^{-19}\text{ J} = 4.14\text{ eV}$. Work function $\phi = E - K_{\max} = 4.14 - 1.2 = 2.94\text{ eV} = 4.7\times10^{-19}\text{ J}$. Longest wavelength is the threshold: $\lambda_0 = \dfrac{hc}{\phi} = \dfrac{1.989\times10^{-25}}{4.7\times10^{-19}} \approx 420\text{ nm}$.