A light-emitting diode (LED) emits light over a narrow range of wavelengths - AQA - A-Level Physics - Question 2 - 2021 - Paper 3

One possible disadvantage of using the fifth-order maximum is that high-order maxima are often less intense and may lead to less precise measurements due to overlapping with higher-order spectral lines.

To find the activation voltage $ext{V}_A$ for the LED $L_R$, we need to extrapolate the linear region of the current-voltage characteristic from Figure 4 until it meets the horizontal axis. The value can be determined to be approximately $1.95 ext{ V}$.

Using the relation:

$V_A = \frac{h c}{e \lambda_p}$

we can rearrange this to find:

$h = \frac{V_A e \lambda_p}{c}$

Inserting the known values:

• $ext{V}_A = 1.95 ext{ V}$
• $e = 1.6 \times 10^{-19} ext{ C}$
• $ext{c} = 3.00 \times 10^8 ext{ m/s}$
• Using $ext{λ}_p$ from the graph (linear interpolation gives an appropriate value, e.g., 500 nm) = $500 \times 10^{-9} ext{ m}$:

$h = \frac{1.95 \times 1.6 \times 10^{-19} \times 500 \times 10^{-9}}{3.00 \times 10^8}$

This calculation yields a value for the Planck constant around $h \approx 6.63 \times 10^{-34} \text{ J.s}$.

Using Ohm's law, we can calculate the minimum value of the resistor $R$. The total voltage from the power supply is $6.10 ext{ V}$.

The maximum current in $L_R$ is $21.0 ext{ mA}$. Therefore:

$V_R = I_{L_R} \times R$

Where:

• $V_R = 6.10 - 2.00 = 4.10 ext{ V}$ (the voltage drop across the resistor)
• Substituting the values gives:

$R = \frac{4.10 ext{ V}}{21.0 \times 10^{-3} ext{ A}} \approx 195 \text{ ohms}$

Thus, the minimum value of $R$ should be 195 ohms.