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A light-emitting diode (LED) emits light over a narrow range of wavelengths - AQA - A-Level Physics - Question 2 - 2021 - Paper 3
Question 2
A light-emitting diode (LED) emits light over a narrow range of wavelengths. These wavelengths are distributed about a peak wavelength $ ext{λ}_p$. Two LEDs $L_G$ a... show full transcript
Worked Solution & Example Answer:A light-emitting diode (LED) emits light over a narrow range of wavelengths - AQA - A-Level Physics - Question 2 - 2021 - Paper 3
Step 1
Determine $N$, the number of lines per metre on the grating.
Answer
To find the number of lines per metre for the grating, we use the diffraction grating formula: where:
- is the distance between adjacent slits (lines per metre),
- is the diffraction angle (76.3),
- is the order of the maximum (5th order),
- is the wavelength (corresponding to ).
First, we rearrange the formula:
Substituting known values will yield:
Assuming is measured/read off from Figure 3, we would have:
Step 2
Suggest one possible disadvantage of using the fifth-order maximum to determine $N$.
Answer
One possible disadvantage is that the fifth-order maximum may result in less accuracy due to several factors such as:
- The higher the order, the more spread out the maxima, which can lead to overlapping or interference from nearby peaks.
- The intensity of the maximum may be lower in higher orders, making it harder to read accurately.
Step 3
Determine, using Figure 4, $V_A$ for $L_R$.
Answer
To determine the activation voltage for , we need to examine Figure 4 to find where the linear part of the characteristic intersects the horizontal axis. By extrapolating this linear region accurately, we can read off the value of . For , it is found to be approximately 1.95 V based on the graph.
Step 4
Deduce a value for the Planck constant based on the data given about the LEDs.
Answer
Using the formula: We can rearrange this to find : Based on the available values from previous parts, substituting the known values will yield:
- For the activation voltage ,
- Assuming is known, substituting these will give a calculated value for .
Step 5
Deduce the minimum value of $R$.
Answer
The minimum resistor value can be calculated using Ohm’s law: where:
- is the supply voltage (6.10 V),
- for (as calculated previously), and
- is the maximum current in (21.0 mA).
Calculating this will ensure the minimum value of does not exceed the overload current.