Photoelectric Effect (Einstein)

Subtract the work function from the photon energy: $E_{\mathrm{kin}} = h\cdot{}f - W_{A}$. The photon energy is $h\cdot{}f$ with Planck constant $h = 6.626\times{}10^{-34} J\cdot{}s$ and the light frequency f in hertz. The work function W_A is the minimum energy to release an electron from the metal, often given in electronvolts. Example sodium with $W_{A} = 2.28 eV$ and $f = 8\times{}10^{14} Hz$: $E_{\mathrm{kin}} =\frac{ (6.626\times{}10^{-34}\cdot{}8\times{}10^{14})}{1.6\times{}10^{-19}} - 2.28 \approx{} 1.03 eV$. Watch for consistent units; work either entirely in joules or entirely in electronvolts. If $h\cdot{}f$ is smaller than W_A, no electrons are emitted at all.

Because the energy per photon depends solely on the frequency, $E = h\cdot{}f$, and an electron always absorbs only a single photon. Higher intensity means more photons per time, but each individual photon has the same energy. Therefore more intense light above the threshold frequency releases more electrons, increasing the photocurrent, but does not change the maximum kinetic energy of the individual electrons. That depends only on f. Exactly this finding contradicted the classical wave theory of light and led Einstein in 1905 to the photon hypothesis, for which he later received the Nobel Prize.

The threshold frequency f_g is the smallest light frequency at which electrons are emitted at all. Exactly at it the photon energy just suffices to cover the work function, so the kinetic energy is zero. From $h\cdot{}f_{g} = W_{A}$ it follows that $f_{g} =\frac{ W_{A}}{h}$. Below this frequency no photoelectrons are emitted, no matter how intense or how long you irradiate. For sodium with $W_{A} = 2.28 eV = 3.65\times{}10^{-19} J$ you get $f_{g} =\frac{ 3.65\times{}10^{-19}}{6.626\times{}10^{-34}} \approx{} 5.5\times{}10^{14} Hz$, which lies in the green range of visible light. Convert W_A from electronvolts to joules before inserting.

The stopping voltage U is the voltage you must apply to just barely stop even the fastest photoelectrons. At the stopping point the electrical work $e\cdot{}U$ equals the maximum kinetic energy: $e\cdot{}U = E_{\mathrm{kin}} = h\cdot{}f - W_{A}$. From this you can calculate $U =\frac{ (h\cdot{}f - W_{A})}{e}$. Measuring U for different frequencies gives a straight line U(f) with slope $\frac{h}{e}$; from it Planck constant is determined experimentally. The intercept yields the threshold frequency. The stopping voltage depends only on the frequency, not on the intensity of the light.

The external photoelectric effect occurs only when the photon energy $h\cdot{}f$ is at least as large as the work function W_A of the material. If the frequency is below the threshold frequency $f_{g} =\frac{ W_{A}}{h}$, the energy of a single photon is not enough to release an electron, and nothing happens, regardless of the light intensity or the exposure time. Red light with low frequency therefore releases no electrons in many metals, while more energetic UV light does. Only above the threshold frequency does the effect begin abruptly, which strikingly demonstrates the quantum nature of light.