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 $ extit{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 N, we use the diffraction grating equation:
dimesextsin(heta)=mextitλ
where:
d is the grating spacing (distance between lines)
m is the order of the maximum (in this case, 5)
heta is the diffraction angle (76.3°)
First, we calculate d:
d = rac{m extit{λ}}{ ext{sin}( heta)}
Assuming a typical wavelength like 650extnm (for LR):
d = rac{5 imes 650 imes 10^{-9}}{ ext{sin}(76.3^ ext{o})}
After calculating d, you can determine N as:
N = rac{1}{d}
Evaluate to get the result.
Step 2
Suggest one possible disadvantage of using the fifth-order maximum to determine $N$.
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Answer
One possible disadvantage is that the fifth-order maximum may be more susceptible to errors due to diffraction limitations, as higher-order maxima can become less distinct and overlap with adjacent orders. This could lead to inaccuracies in determining the exact angle and therefore the value of N.
Step 3
Determine, using Figure 4, $V_A$ for $L_R$.
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Answer
From Figure 4, extrapolate the linear region of the LR characteristic to find the intersection with the horizontal axis. This point represents the activation voltage VA. Read the voltage value directly from the graph to get 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 derived relationship:
V_A = rac{hc}{e extit{λ}_p}
For LG, we have:
VA=2.00extV
extitλp=520extnm=520imes10−9extm
Rearranging gives:
h = rac{V_A imes e imes extit{λ}_p}{c}
Substituting the known values (e=1.6imes10−19extC, c=3.0imes108extm/s) will yield a calculated value for the Planck constant.
Step 5
Deduce the minimum value of $R$.
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Answer
Using Ohm's law, the voltage across the resistor R can be evaluated using:
VR=extemf−VA
Substituting values gives:
VR=6.10extV−VA
The current through R is 21.0 mA, thus:
R = rac{V_R}{I}
Calculate R using this formula to find the minimum resistance.
