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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

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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 find the number of lines per metre (N), we can use the grating equation: dsin⁡(θ)=nλd \sin(\theta) = n\lambda where:

  • d is the distance between adjacent slits (the grating spacing),
  • (\theta = 76.3^{\circ}),
  • n is the order of the maximum (n=5).

First, we need to find (\lambda_p) from the graph in Figure 3. Assuming (\lambda_p) is around 650 nm for red light, we convert it to meters: (\lambda_p = 650 \times 10^{-9} , m).

Now substituting the values: d=nλsin⁡(θ)=5×650×10−9sin⁡(76.3∘)≈1.37×10−7 md = \frac{n\lambda}{\sin(\theta)} = \frac{5 \times 650 \times 10^{-9}}{\sin(76.3^{\circ})} \approx 1.37 \times 10^{-7} \, m

The number of lines per metre is: N=1d≈7.305×106 m−1.N = \frac{1}{d} \approx 7.305 \times 10^{6} \, m^{-1}.

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 is more susceptible to errors in measurement due to diffraction spreading. As the order increases, the maxima may become closer together, making them harder to distinguish and measure accurately. Additionally, the intensity of the light may decrease, leading to weaker signals.

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 current-voltage characteristic for LR_R to find the intersection with the horizontal axis. This intersection represents VaV_a. By reading off the graph, we find:

Va=2.1 VV_a = 2.1 \text{ V}.

Step 4

Deduce a value for the Planck constant based on the data given about the LEDs.

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Answer

Using the formula: Va=hceλpV_a = \frac{hc}{e\text{λ}_p}. We can rearrange it to solve for h: h=Vaeλpch = \frac{V_a e \text{λ}_p}{c}. Given:

  • Va=2.0 VV_a = 2.0 \text{ V} for LG_G,
  • λp\text{λ}_p is approximately 650×10−9 m650 \times 10^{-9} \text{ m},
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C},
  • c=3.0×108 m/sc = 3.0 \times 10^{8} \text{ m/s}, yielding:

h \approx \frac{2.0 \times 1.6 \times 10^{-19} \times 650 \times 10^{-9}}{3.0 \times 10^{8}}$$$$\approx 4.14 \times 10^{-34} \text{ Js}.

Step 5

Deduce the minimum value of R.

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Answer

Using Ohm's law, we can determine the minimum resistance R. Given the total voltage, V=6.10 VV = 6.10 \text{ V}, and the maximum current through LR_R, I=21.0 mA=0.021 AI = 21.0 \text{ mA} = 0.021 \text{ A}, we find:

R=VI=6.100.021≈290.48 Ω.R = \frac{V}{I} = \frac{6.10}{0.021} \approx 290.48 \, \Omega. Therefore, the minimum value of R must be at least 290.48 Ω290.48 \, \Omega.

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