Photons and the Photoelectric Effect

Wavelength, Frequency and the Spectrum described light as a wave. Around 1900, two puzzles showed that this is only half the story: the colors glowing from hot objects, and the way light knocks electrons out of metals. Both are explained if light's energy comes in packets, called photons.

Key idea

Each photon carries energy $$E = h\nu = \frac{hc}{\lambda},$$ where $h = 6.626 \times 10^{-34}\ \text{J·s}$ is Planck's constant. Higher frequency (shorter wavelength) means more energy per photon. Brighter light means more photons, not more energetic ones.

The photoelectric effect

Shine light on a clean metal and electrons can fly off. The surprising facts: below a certain threshold frequency, no electrons leave, however bright the light. Above it, electrons leave at once, even in dim light, and their maximum kinetic energy grows with the frequency but not with the brightness.

Why it works

In 1905 Einstein explained this with photons. One photon gives all its energy to one electron. The electron needs a minimum energy $\Phi$ (the work function, set by how tightly the metal holds its electrons) to escape. If $h\nu < \Phi$, no single photon is enough, and in ordinary light piling on more photons does not help, because an electron does not save up energy from several photons. If $h\nu > \Phi$, the leftover energy becomes motion: $KE_\text{max} = h\nu - \Phi$. Brighter light sends more photons, so more electrons leave, each with the same top speed.

Sodium's work function is $2.46\ \text{eV}$ ($3.94 \times 10^{-19}\ \text{J}$), so its threshold is about 504 nm. Bright red light frees nothing; dim blue light frees electrons at once.

Worked example: Energy of one photon and of a mole of photons

Find the energy of a photon of $500\ \text{nm}$ light, and of a mole of such photons.

$$E = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34}\ \text{J·s})(3.00 \times 10^{8}\ \text{m/s})}{5.00 \times 10^{-7}\ \text{m}} = 3.98 \times 10^{-19}\ \text{J}$$
Per mole: $$3.98 \times 10^{-19}\ \text{J} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 2.39 \times 10^{5}\ \text{J/mol} = 239\ \text{kJ/mol}$$
That is the size of a chemical bond's energy, which is why visible and UV light can drive chemistry.

Answer: $3.98 \times 10^{-19}\ \text{J}$ per photon; $239\ \text{kJ/mol}$.

Worked example: Will electrons leave?

Sodium's work function is $3.94 \times 10^{-19}\ \text{J}$. Light of $400\ \text{nm}$ shines on it. Do electrons leave, and with what maximum kinetic energy?

Photon energy: $$E = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{4.00 \times 10^{-7}} = 4.97 \times 10^{-19}\ \text{J}$$
That is more than $\Phi$, so yes.
$$KE_\text{max} = 4.97 \times 10^{-19} - 3.94 \times 10^{-19} = 1.03 \times 10^{-19}\ \text{J}$$

Answer: Yes; $1.03 \times 10^{-19}\ \text{J}$.

Worked example: Counting photons

A green laser pulse at $532\ \text{nm}$ carries $1.00\ \text{mJ}$ of energy. How many photons does it hold?

One photon: $$E = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{5.32 \times 10^{-7}} = 3.74 \times 10^{-19}\ \text{J}$$
Photons $= \dfrac{1.00 \times 10^{-3}\ \text{J}}{3.74 \times 10^{-19}\ \text{J}} = 2.68 \times 10^{15}$.

Answer: $2.68 \times 10^{15}$ photons.

Common mistake

Thinking brighter light gives each photon more energy, so a bright red beam should free electrons that a dim violet one can't. Brightness is the number of photons; a red photon has the same small energy in a dim beam or a bright one. Check: compare $h\nu$ of a single photon with the work function. If one photon falls short, a trillion of them still free nothing.

Photons and the Photoelectric Effect

The photoelectric effect

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