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 $\lambda_p$.
Two LEDs L_G an... 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) on the grating, we can use the diffraction formula:
dsin(θ)=nλ
where:
d is the distance between the grating lines (the grating spacing),
\theta is the angle of diffraction (76.3∘),
n is the order of the maximum (n = 5 for the fifth-order maximum),
\lambda is the wavelength associated with the maximum intensity.
Rearranging the formula gives:
N=d1=λsin(θ)n
Assuming \lambda_p is determined from the graph (e.g., if \lambda_p = 650 nm = 650 \times 10^{-9} m), we can calculate:
N=(650×10−9)sin(76.3∘)5≈3.06×103 lines/m
Step 2
Suggest one possible disadvantage of using the fifth-order maximum to determine N.
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Answer
One possible disadvantage of using the fifth-order maximum is that it might be blurred or less distinct compared to lower-order maxima. The intensity of the fifth-order maximum is generally weaker, potentially making it harder to accurately measure the angle of diffraction. This could lead to larger errors in calculating N.
Step 3
Determine, using Figure 4, V_A for L_R.
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Answer
To determine V_A for L_R, we need to extrapolate the linear region of the I-V characteristic curve until it intersects the horizontal axis, which indicates the activation voltage. Based on Figure 4, this extrapolated value appears to meet the horizontal axis at approximately 1.92 V.
Step 4
Deduce a value for the Planck constant based on the data given about the LEDs.
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Answer
Using the formula:
VA=eλphc
we can rearrange to find Planck's constant (h):
h=VA⋅e⋅λp/c
Assuming e = 1.602×10−19C and taking both V_A for L_R (approximately 1.92 V) and \lambda_p = 650 \times 10^{-9} m,
we substitute the values:
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Answer
Using Ohm's Law and the given power supply conditions:
V=I⋅R
where V = 6.10 V and I should not exceed 21.0 mA (0.021 A).
Thus,
R≥IV=0.0216.10≈290.48Ω
Thus, the minimum value of R is approximately 290.5 Ω.
