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 it may lead to reduced accuracy in measurements due to a wider peak for the maximum. This affects the precision of locating the exact maximum and can lead to increased systematic error in the result.

To find the activation voltage ($V_A$) for $L_R$, refer to the linear part of the current-voltage characteristic in Figure 4. Extrapolating this linear part to the horizontal axis gives the value for $V_A$. From the graph, you would read $V_A \approx 2.25 V$.

Using the equation for the activation voltage: $V_A = \frac{hc}{e \lambda_p}$ We can rearrange to find: $h = \frac{V_A \cdot e \cdot \lambda_p}{c}$ Substituting:

• $V_A = 2.00 V$,
• Use $e \approx 1.6 \times 10^{-19} C$,
• $\lambda_p = 650 \times 10^{-9} m$,
• $c = 3.00 \times 10^{8} m/s$. This gives: $h = \frac{(2.00)(1.6 \times 10^{-19})(650 \times 10^{-9})}{3.00 \times 10^{8}} \approx 6.63 \times 10^{-34} J.s$

To find the minimum value of the resistor $R$, use Ohm's Law, where: $V = IR$ Rearranging gives: $R = \frac{V}{I}$ Here, $V = 6.10 V$ (emf of the power supply) and $I = 21.0 mA = 0.021 A$. Substituting the values: $R = \frac{6.10}{0.021} \approx 290.48 \Omega$ Thus, the minimum value of $R$ should be rounded up to $291 \Omega$ to ensure the current does not exceed $21.0 mA$.