LED Turn-On Voltage and Planck’s Constant
A student determines Planck’s constant using the turn-on voltage of five LEDs. The LEDs have peak wavelengths 470, 525, 590, 630 and 660 nm. Each is connected in series with a 330 Ω resistor and a variable d.c. supply. The student increases the voltage slowly and records the supply voltage at which a current of 0.10 mA is first measured through the LED; the voltage across the LED itself is not measured directly. The readings are 2.62, 2.35, 2.08, 1.96 and 1.87 V. The student plots these voltages against 1/λ and obtains a straight line of gradient 1.22×10−6 V m and intercept −0.32 V. A digital multimeter with 10 MΩ input resistance is used to check the supply voltage; its reading is 0.03 V lower than a calibrated meter at 2.5 V. The room temperature changes by 3 K during the experiment.
1. Use the gradient to calculate a value of Planck’s constant and compare it with 6.63×10−34 J s. Explain why the intercept is negative and whether it should be included in the calculation.
Read the worked answer
For photon energy, eV ≈ hc/λ, so V = (hc/e)(1/λ) + intercept. The gradient is hc/e, hence
h = e × gradient / c = 1.602×10−19 × 1.22×10−6 / 3.00×10⁸ ≈ 6.52×10−34 J s.
This is about 1.7% below the accepted 6.63×10−34 J s. The negative intercept (−0.32 V) represents contact potential, series resistance, and the fact that the LED does not turn on exactly at the photon-energy threshold; some energy is lost before detectable emission. A constant offset changes the intercept but not the gradient, so the intercept should not be forced through zero. The slope is the quantity that gives h.
2. The student records the supply voltage, not the LED voltage. Estimate the voltage dropped across the 330 Ω resistor at the 0.10 mA criterion and decide whether this correction changes h. What if the current criterion were 5 mA?
Read the worked answer
V_R = IR = 0.10×10−3 × 330 = 0.033 V. This is a nearly constant offset added to every LED voltage, so it shifts the intercept by about 0.03 V but leaves the gradient, and therefore h, essentially unchanged.
At 5 mA, V_R = 5×10−3 × 330 = 1.65 V. This is a much larger offset, but if the same current criterion is used for every LED it is still approximately constant and again mainly affects the intercept. However, at 5 mA the LEDs heat more and their dynamic resistance differs, so the constant-offset assumption becomes poorer. The multimeter’s 0.03 V error is also a constant offset at these readings and does not explain a gradient error.
3. The 470 nm LED heats by 6 K during measurement and its wavelength increases by 0.05 nm K−1. Estimate whether this could explain the discrepancy in h.
Read the worked answer
Wavelength shift: Δλ = 0.05 × 6 = 0.30 nm. For a LED, a rough forward-voltage temperature coefficient is about −2 mV K−1, so 6 K gives about −12 mV on the 470 nm point. The slope is determined mainly by the voltage difference between the shortest and longest wavelengths: Δ(1/λ) ≈ 1/470 nm − 1/660 nm = 6.13×10⁵ m−1. A 12 mV error at one end changes the gradient by about 0.012/6.13×10⁵ ≈ 2.0×10−8 V m, or about 1.6% of 1.22×10−6 V m. That is the right order of magnitude to explain the observed 1.7% low value of h. The 470 nm point should therefore be treated with suspicion; measuring at a fixed low current and allowing the LED to reach thermal equilibrium would test this.