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

One disadvantage of using the fifth-order maximum to determine $N$ is that the fifth-order maximum may be weaker and less distinct compared to lower order maxima, making it harder to measure accurately. Additionally, higher-order maximum could lead to overlapping wavelengths, which can introduce errors in the measurement.

To find the activation voltage $V_A$ for $L_R$, we need to extrapolate the linear portion of its current-voltage characteristic until it intersects the horizontal axis. By checking the graph, it can be determined that $V_A$ for $L_R$ is approximately 2.2 V.

Using the relationship $V_A = \frac{hc}{e\tau_p}$ and substituting $V_A$ for $L_G$, we can rearrange the equation to find $h$: $h = \frac{V_A e \tau_p}{c}$ Substituting known values (where $e = 1.60 \times 10^{-19} \text{ C}, c = 3.00 \times 10^8 \text{ m/s}$) and calculating will yield a value for Planck's constant within an acceptable range.

To find the minimum resistance $R$, we can use Ohm’s law and the formula for power: $V = IR$ Where $V = 6.10 V$ and $I$ is the maximum current of $21.0 mA$:

Rearranging gives: $R = \frac{V}{I} = \frac{6.10}{0.021}$ Calculating this results in a minimum value of approximately $290.48 \Omega$. Therefore, the minimum value of resistance $R$ must be at least $290.48 \Omega$.