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 $L_G$ ... 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.
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Answer
To find the number of lines per metre N, we can use the diffraction formula:
dsin(θ)=nλ
where:
d is the distance between the lines of the grating,
heta is the angle of the maximum ( ext{76.3}^\text{o}),
n is the order of the maximum (5 for fifth order),
λ is the wavelength.
From the given heta, we can calculate N:
Convert angle to radians: 76.3exto=18076.3×π.
Use the equation:
N=d1=λsin(θ)n
This results in:
N≈3.06×103 m−1
Step 2
Suggest one possible disadvantage of using the fifth-order maximum to determine $N$.
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Answer
One disadvantage is that the fifth-order maximum might be difficult to observe accurately due to its dispersion and reduced intensity compared to lower orders. This can lead to measurement errors when interpolating results.
Step 3
Determine, using Figure 4, $V_A$ for $L_R$.
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Answer
To find the activation voltage VA for LR, examine the graph in Figure 4:
Locate the linear portion of the I−V characteristic for LR.
Extrapolate the linear section to find the intersection with the horizontal axis.
This reading will give the value of VA.
Step 4
Deduce a value for the Planck constant based on the data given about the LEDs.
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Answer
Using the relationship VA=eλphc:
Rearranging for h, we have:
h=cVAeλp.
Substitute in values for VA, e, and c.
This will yield a calculated value for the Planck constant.
Step 5
Deduce the minimum value of $R$.
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Answer
From the circuit, we have:
Supply voltage: 6.10extV
Current through LR: must not exceed 21.0extmA.
Using Ohm's law:
R=IV=21.0×10−3extA6.10extV≈290.48extΩ.
Therefore, the minimum value of R is approximately 290.48extΩ.
