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 and... 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
0 2 . 1 Determine N, the number of lines per metre on the grating.
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Answer
To find the number of lines per meter (N) on the grating, we can use the grating equation:
dsinθ=nλ
where:
d is the distance between grating lines (grating spacing),
heta is the diffraction angle,
n is the order of the maximum (in this case, n = 5),
extλ is the wavelength.
Given:
heta=76.3exto and
we assume a typical wavelength for red light extλp≈650 nm=650×10−9extm.
We can rearrange to find:
d=sinθnλ
Substituting in the known values gives:
d=sin(76.3)5×650×10−9≈3.06×10−6 m
The number of lines per meter is:
N=d1=3.06×10−61≈3.26×105 lines/m.
Step 2
0 2 . 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 to determine N is that higher-order maxima are typically less intense and may be less distinct. As a result, it can be more difficult to accurately identify and measure the position of the fifth-order maximum, which could lead to errors in the calculated value for N.
Step 3
0 2 . 3 Determine, using Figure 4, $V_A$ for L_R.
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Answer
To determine the activation voltage VA for L_R, we need to extrapolate the linear region of the current-voltage characteristic from Figure 4 and read the point where it meets the horizontal axis. From the graph, we find that the activation voltage VA for L_R is approximately 1.85 V.
Step 4
0 2 . 4 Deduce a value for the Planck constant based on the data given about the LEDs.
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Answer
Using the equation:
VA=eλphc
For L_G, we can rearrange to find:
h=cVAeλp
Where:
e≈1.60×10−19C (charge of an electron),
c≈3.00×108m/s (speed of light), and
extλp≈650nm=650×10−9m.
Substituting these values gives:
h≈3.00×108(2.00)(1.60×10−19)(650×10−9)≈6.63×10−34Js.
Step 5
0 2 . 5 Deduce the minimum value of R.
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Answer
Using Ohm's Law, the minimum value of the resistor R can be determined as follows:
Given the emf from the power supply is 6.10 V and the maximum current is 21.0 mA (or 0.021 A), we can use the equation:
V=IR
Rearranging gives:
R=IV=0.0216.10≈290.48Ω.
Thus, the minimum value of R should be at least 290.48 Ω.
