Einstein's photoelectric equation.

• In 1905 Einstein proposed that radiation energy comes in discrete units, quanta, each of energy hν, where h is Planck's constant and ν the frequency. Emission is not the slow soaking-up of energy from a wave.
• An electron absorbs one whole quantum hν. If that exceeds the work function φ₀, the electron escapes, and the most it can carry away is Kmax = hν − φ₀ (Eq. 11.2). Electrons bound more tightly come out with less than this.
• Kmax depends linearly on ν and not on intensity, because one electron absorbs one quantum; intensity only fixes how many quanta arrive per unit area per second.
• Since Kmax cannot be negative, emission needs hν > φ₀, that is ν > ν₀ with ν₀ = φ₀/h (Eq. 11.3). A metal with a larger work function has a higher threshold frequency, and below ν₀ nothing comes out, however intense the light or however long it shines.
• More quanta per second means more electrons absorbing them, so for ν > ν₀ the photocurrent is proportional to intensity. And since each absorption is a single, instantaneous event, dim light causes no delay: it only means fewer electrons take part.
• With Kmax = eV₀ the equation becomes eV₀ = hν − φ₀, or V₀ = (h/e)ν − φ₀/e for ν ≥ ν₀ (Eq. 11.4). The V₀–ν graph is a straight line of slope h/e, the same for every material.
• Millikan, between 1906 and 1916, set out to disprove the equation. He measured the slope of the line for sodium and, using the known e, found a value of h close to Planck's constant (6.626 × 10⁻³⁴ J s) obtained in an entirely different context. He ended up confirming the equation for several alkali metals over a wide range of frequencies.
• Example 11.2 (caesium, φ₀ = 2.14 eV): 2.14 eV is 3.42 × 10⁻¹⁹ J, and dividing by h = 6.63 × 10⁻³⁴ J s gives ν₀ = 5.16 × 10¹⁴ Hz. A 0.60 V stopping potential needs hν = 0.60 + 2.14 = 2.74 eV, so λ = hc/hν = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(2.74 × 1.6 × 10⁻¹⁹ J) = 454 nm.
• Exercise 11.5: a V₀–ν slope of 4.12 × 10⁻¹⁵ V s gives h = e × slope = 1.6 × 10⁻¹⁹ × 4.12 × 10⁻¹⁵ = 6.59 × 10⁻³⁴ J s. Exercise 11.6: with ν₀ = 3.3 × 10¹⁴ Hz and ν = 8.2 × 10¹⁴ Hz, V₀ = h(ν − ν₀)/e = 6.63 × 10⁻³⁴ × 4.9 × 10¹⁴/(1.6 × 10⁻¹⁹) = 2.0 V.
• Exercise 11.7: a photon of wavelength 330 nm carries hc/λ = 6.03 × 10⁻¹⁹ J = 3.77 eV, less than a work function of 4.2 eV, so that metal gives no photoelectric emission. Exercise 11.9: 488 nm light carries 2.55 eV; with a stopping potential of 0.38 V the work function is 2.55 − 0.38 = 2.16 eV.

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