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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 \(\lambda_p\). 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.
Answer
To find the number of lines per metre, we can use the diffraction formula:
Where:
- (d) is the distance between grating lines (in metres)
- (\theta) is the diffraction angle (76.3°)
- (n) is the order of the maximum (5th order)
- (\lambda) is the wavelength corresponding to (\lambda_p)
First, we convert the angle to radians for calculations: [ \theta = 76.3^{\circ} \approx 1.33 \text{ radians} ]
Now we can rearrange the formula to calculate (d) as:
If we take (\lambda_p) as 650 nm (0.00065 m) for calculations: [ d = \frac{5 \times 0.00065}{\sin(76.3^{\circ})} \approx 0.000104 m ]
The number of lines per metre (N ) is given by: [ N = \frac{1}{d} \approx \frac{1}{0.000104} \approx 9615 \text{ lines/m} ]
Step 2
Suggest one possible disadvantage of using the fifth-order maximum to determine \(N\).
Answer
One potential disadvantage is that the fifth-order maximum is more susceptible to measurement errors due to reduced intensity and potential overlap with other maxima. This can cause inaccuracies in determining the exact angle for the fifth-order maximum, leading to errors in the calculated value of (N). Additionally, higher orders can be less distinct, making them harder to measure accurately.
Step 3
Determine, using Figure 4, \(V_A\) for \(L_R\).
Answer
To find (V_A) for (L_R), we need to extrapolate the linear region of the current-voltage characteristic in Figure 4 to the x-axis. This involves drawing a straight line through the linear portion of the curve until it intersects the x-axis. The x-coordinate at this intersection gives (V_A). After analyzing Figure 4, we find that (V_A \approx 1.80 V) for (L_R).
Step 4
Deduce a value for the Planck constant based on the data given about the LEDs.
Answer
Using the equation for the activation voltage:
We can rearrange this to solve for the Planck constant (h):
Substituting values:
- (V_A \approx 2.00 V) (for (L_G))
- Charge of an electron, (e \approx 1.6 \times 10^{-19} C)
- Speed of light, (c \approx 3.0 \times 10^8 m/s)
- Wavelength, (\lambda_p \approx 650 \times 10^{-9} m)
Now plug in the values:
Step 5
Deduce the minimum value of \(R\).
Answer
We use Ohm's law, which states:
Given:
- Total voltage of the supply is 6.10 V.
- The maximum current (I) in (L_R) should not exceed 21.0 mA (0.021 A).
Thus, we rearrange Ohm's law:
Substituting the values:
Therefore, the minimum resistance (R) must be at least 290.48 Ω (rounded to the nearest standard resistor value, 330 Ω could be used).