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.
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Answer
To determine N, we can use the formula for the diffraction maxima:
dimesextsin(heta)=mextλ
where:
d is the distance between the grating lines (1/N),
m is the order of the maximum (which is 5 in this case),
θ is the angle of diffraction (76.3exto),
extλ is the wavelength of light at the peak intensity.
Rearranging gives:
N=dm=λimessin(θ)m
By taking the corresponding wavelength extλp, which is readable off the peak in Figure 3, substituting values, we find:
N=3.06imes103m−1
Step 2
Suggest one possible disadvantage of using the fifth-order maximum to determine N.
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One possible disadvantage could be that the fifth-order maximum is wider compared to lower order maxima, making it less precise to measure accurately. Furthermore, higher orders may be affected by diffraction limits and may not clearly define a maximum due to the overlapping of light.
Step 3
Determine, using Figure 4, $V_A$ for L$_r$.
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Answer
From the characteristics shown in Figure 4, visually extrapolate the linear region for Lr. The intersection with the horizontal axis indicates the activation voltage VA. From the graph, we find:
VA=2.1extV.
Step 4
Deduce a value for the Planck constant based on the data given about the LEDs.
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Using the activation voltage formula:
VA=eλphc
Substituting available values, particularly using VA=2.1 V for Lr, we derive:
h=cVAimeseimesextλp
Calculating for an appropriate range of wavelengths from Figure 3 will give a mean value for the Planck constant, which is approximately 6.63imes10−34extJs.
Step 5
Deduce the minimum value of R.
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Using Ohm's law:
R=IV
Where:
V is the supply voltage, 6.10 V,
I is the current through Lr, which must not exceed 21.0 mA or 0.021 A.
Substituting these values gives:
R=0.0216.10≈290.48extΩ
Therefore, the minimum value of R must be at least 290 Ω.
