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

To determine $V_{A}$ for L$_{R}$, we need to extrapolate the linear region of the current-voltage characteristic for L$_{R}$ in Figure 4. This is done by extending the straight line region of the curve until it intersects the horizontal axis (indicating where the current is zero). The voltage corresponding to this point on the horizontal axis is the activation voltage $V_{A}$ for L$_{R}$. After analyzing the graph, suppose we find that $V_{A} = 1.75 ext{ V}$.

Using the formula V_{A} = rac{hc}{e ext{λ}_{p}}, we rearrange to solve for $h$:

h = rac{V_{A} imes e imes ext{λ}_{p}}{c}

Where:

• $e$ is the elementary charge, which is approximately $1.6 imes 10^{-19} C$.
• $c$ is the speed of light, approximately $3.0 imes 10^{8} m/s$.
• Suppose $ext{λ}_{p} = 650 imes 10^{-9} m$ and $V_{A} = 1.75 V$.

Substituting these values into the equation will give us the Planck constant. Calculate:

h = rac{1.75 imes (1.6 imes 10^{-19}) imes (650 imes 10^{-9})}{3.0 imes 10^{8}} Evaluate this expression to find the value of $h$.

Using Ohm’s law, the relationship between voltage, current, and resistance is given by: $V = IR,$ where $V$ is the total voltage provided by the power supply, $I$ is the current flowing through L$_{R}$, and $R$ is the resistance.

The total voltage supplied is 6.10 V. The maximum current in L$_{R}$ should not exceed 21.0 mA (or $21.0 imes 10^{-3} A$). To maintain the safety of the LED, we will use:

R = rac{V}{I} = rac{6.10}{21.0 imes 10^{-3}}

Calculating this will provide the minimum resistance $R$. By solving this equation, we determine the value of $R$.