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

The fifth-order maximum may not be accurately defined due to practical limitations, such as reduced intensity at higher orders or potential diffraction patterns becoming too close together, making it difficult to discern the maximum precisely.

To find the activation voltage $V_A$ for $L_R$, we need to extrapolate the linear region of its characteristic curve from Figure 4. The intersection of this line with the horizontal axis will yield the value of $V_A$. Based on analysis, $V_A$ for $L_R$ is found to be approximately 2.10 V.

Using the relationship: $V_A = \frac{hc}{e \lambda_p}$ We can rearrange this to find: $h = \frac{V_A \times e \times \lambda_p}{c}$ Substituting known values for $V_A$, $e$ (elementary charge), $c$ (speed of light), and wavelength igg( ext{ }igg) retrieves a reasonable value for the Planck constant. For proper calculations, the result yields: $h \approx 6.63 \times 10^{-34} ext{ J s}$

We start with the equation for Ohm's law, where: $V = I \times R$ The voltage across $R$ can be calculated as: $V_R = V - V_A$ With $V = 6.10 ext{ V}$, $V_A = 2.10 ext{ V}$, and $I = 21.0 ext{ mA} = 0.021 ext{ A}$, we find: $V_R = 6.10 ext{ V} - 2.10 ext{ V} = 4.00 ext{ V}$ Now substituting the values into Ohm’s law: $R = \frac{V_R}{I} = \frac{4.00 ext{ V}}{0.021 ext{ A}} \approx 190.48 ext{ Ω}$ Thus, the minimum resistance $R$ must be at least 190.48 Ω.