Simulation · Physics · Class 12
Einstein's photoelectric equation
From the lesson Einstein's photoelectric equation in Dual Nature of Radiation and Matter. Change the values and watch what happens.
The idea behind it
NCERT §11.6
- 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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