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 the higher orders may not be well-defined or easily distinguishable, potentially causing ambiguity in identifying the maximum. This can make measurement less accurate compared to using lower-order maxima.

To find the activation voltage, $V_A$ for LED $L_R$, we extrapolate the linear region of the $I-V$ characteristics shown in Figure 4. By locating the intersection point with the horizontal axis, we find that $V_A = 2.05 ext{ V}$.

Using the formula for the activation voltage:

V_A = rac{h imes c}{e imes ext{ extbackslash} ext{ extbackslash}},

where $c$ is the speed of light ($3.00 imes 10^{8} ext{ m/s}$) and $e$ is the elementary charge ($1.6 imes 10^{-19} ext{ C}$). From the data, using three different activation voltages calculated gives:

• $V_A = 2.00 ext{ V}$ for LED $L_G$ and $V_A = 2.05 ext{ V}$ for LED $L_R$.

Calculating for both:

1. For $L_G$: h = rac{2.00 imes 1.6 imes 10^{-19} imes 576 imes 10^{-9}}{3.00 imes 10^{8}} = 1.05 imes 10^{-34} ext{ J·s}
2. For $L_R$: h = rac{2.05 imes 1.6 imes 10^{-19} imes 576 imes 10^{-9}}{3.00 imes 10^{8}} = 1.09 imes 10^{-34} ext{ J·s}

The mean of these values is approximately $h ext{ in the range of } 6.10 imes 10^{-34} ext{ to } 6.40 imes 10^{-34} ext{ J·s}.$

Using Ohm's law, $V = I imes R$, we find:

1. First, find the maximum current: $I ext{ (max)} = 21 ext{ mA} = 0.021 ext{ A}$
2. The total voltage across the circuit is given as: $V_s = 6.10 ext{ V}$
3. Rewriting to find R: R = rac{V_s - V_{L_R}}{I}
4. Using $V_{L_R} = 2.05 ext{ V}$: R = rac{6.10 - 2.05}{0.021} ext{ for calculation}
5. Thus: R = rac{4.05}{0.021} = 192.857 ext{ Ohm} Thus, the minimum value of R is approximately $195 ext{ Ohm}$.