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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
0 2 . 1 Determine $N$, the number of lines per metre on the grating.
Answer
To determine the number of lines per metre () on the grating, we can use the diffraction formula: where is the distance between the grating lines and is the order of the maximum.
Given:
- (fifth-order maximum)
- , which needs to be read off from Figure 3; let's assume it is around 650 nm (0.65 µm).
We can rearrange this formula to find : N = rac{1}{d} = rac{m}{ ext{λ} imes ext{sin}( heta)} Substituting in the values gives: N = rac{5}{(650 imes 10^{-9}) imes ext{sin}(76.3^ ext{o})} Calculating this will provide in lines per metre.
Step 2
0 2 . 2 Suggest one possible disadvantage of using the fifth-order maximum to determine $N$.
Answer
One possible disadvantage is that the fifth-order maximum may be less clear or less intense compared to lower orders. This could make it more difficult to accurately identify and measure the peak of the maximum, leading to potential errors in the calculation of .
Step 3
0 2 . 3 Determine, using Figure 4, $V_A$ for $L_R$.
Answer
To find for , we can extrapolate the linear region of the characteristic curve for until it intersects the horizontal axis. Since for is given as 2.00 V, we ascertain for by locating the corresponding intersection point on the graph.
Step 4
0 2 . 4 Deduce a value for the Planck constant based on the data given about the LEDs.
Answer
Using the relationship: V_A = rac{h c}{e ext{λ}_p} We can express Planck's constant as: h = rac{V_A e ext{λ}_p}{c} We know:
- The values for and (use the peak wavelength from Figure 3). After substituting these values, we should arrive at a numerical value for .
Step 5
0 2 . 5 Deduce the minimum value of $R$.
Answer
Given the total voltage from the power supply is 6.10 V and the maximum current in is 21.0 mA, we can apply Ohm's Law: Rearranging gives: R = rac{V}{I} = rac{6.10}{0.021} This calculation will yield the minimum resistance , ensuring that the current does not exceed the specified limit.