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 can be harder to accurately measure the angle due to increased diffraction spread, which can lead to measurement errors. As higher orders can often result in more pronounced overlap of adjacent maxima, leading to difficulties in isolation.

To determine the activation voltage $V_A$ for $L_R$, we extrapolate the linear portion of the current-voltage characteristic curve intersecting the horizontal axis. From Figure 4, we see this intersection occurs at approximately 2.1 V.

From the equation $V_A = \frac{hc}{e \tau_p}$, rearranging gives: $h = \frac{V_A e \tau_p}{c}$ Substituting the known values, where $V_A$ for $L_G$ is 2.00 V, $e = 1.6 \times 10^{-19} C$, $c = 3.00 \times 10^8 m/s$, and using an approximate $au_p$ value of 650 nm (which is $6.50 \times 10^{-7} m$) leads to: $h = \frac{(2.00)(1.6 \times 10^{-19})(6.50 \times 10^{-7})}{(3.00 \times 10^8)}$ Calculating this yields a Planck constant value in a range consistent with known values.

Using Ohm's law and the principle of total voltage in a series circuit, we have: $V = I R + V_A$ Where the total voltage supplied is 6.10 V and $I$ must be less than or equal to 21.0 mA. Substituting gives: $6.10 = (21.0 \times 10^{-3})R + 2.00$ Rearranging to solve for $R$ provides: $R = \frac{6.10 - 2.00}{21.0 \times 10^{-3}}$ Thus gives a minimum resistance of about 195 Ω.