Problem — Deriving h from photoelectric data

Problem

In a photoelectric effect experiment you get two data points: $\lambda_1 = 450$ nm gives $V_{a,1} = 0{,}50$ V; $\lambda_2 = 300$ nm gives $V_{a,2} = 1{,}88$ V. Estimate Planck’s constant by eliminating the work function $W$ through subtraction of the two Einstein equations.

Solution

Step 1 — Einstein’s equation. The maximum kinetic energy of the electrons, measured by the stopping potential $V_a$, is

$eV_a = h\nu - W = \frac{hc}{\lambda} - W$

Step 2 — System of two equations.

$eV_{a,1} = \frac{hc}{\lambda_1} - W, \qquad eV_{a,2} = \frac{hc}{\lambda_2} - W$

Step 3 — Subtraction (eliminates $W$).

$e\,(V_{a,2} - V_{a,1}) = hc\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)$

from which

$h = \frac{e\,(V_{a,2} - V_{a,1})}{c\left(\dfrac{1}{\lambda_2} - \dfrac{1}{\lambda_1}\right)}$

Step 4 — Numbers. $V_{a,2} - V_{a,1} = 1{,}88 - 0{,}50 = 1{,}38$ V.

$\frac{1}{\lambda_2} - \frac{1}{\lambda_1} = \frac{1}{300\cdot 10^{-9}} - \frac{1}{450\cdot 10^{-9}} = (3{,}33 - 2{,}22)\cdot 10^{6} = 1{,}11\cdot 10^{6}\;\text{m}^{-1}$

$h = \frac{1{,}6\cdot 10^{-19}\cdot 1{,}38}{3{,}0\cdot 10^{8}\cdot 1{,}11\cdot 10^{6}} = \frac{2{,}21\cdot 10^{-19}}{3{,}33\cdot 10^{14}} \approx 6{,}6\cdot 10^{-34}\;\text{J·s}$

$\ev{h \approx 6{,}6\cdot 10^{-34}\;\text{J·s}}$

Consistent with the known value $h = 6{,}63\cdot 10^{-34}$ J·s. Note that the subtraction eliminates the work function $W$, which depends on the metal: the slope $h/e$ of the $K_\text{max}$ vs $\nu$ graph is instead universal.

Collegamenti

Argomenti: Quantum physics Concetti: Photoelectric effect · Photon Competenze: Symbolic setup · Unit conversion