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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 $ ho_p$. Two LEDs L_G and L_R... 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 NN, we use the diffraction formula for a grating:

dsin⁡θ=mλd \sin \theta = m \lambda

where:

  • dd is the distance between adjacent slits (lines),
  • mm is the order of the maximum (in this case, m=5m = 5),
  • heta=76.3∘ heta = 76.3^\circ is the angle of diffraction, and
  • hop ho_p is the wavelength of the light.

From Figure 3, we can assume hop ho_p is around 650 nm (for red light). Converting this to metres gives: λ=650×10−9 m\lambda = 650 \times 10^{-9} \text{ m}

Solving for dd:

d=λsin⁡θ=650×10−9sin⁡(76.3∘)≈8.32×10−9 md = \frac{\lambda}{\sin \theta} = \frac{650 \times 10^{-9}}{\sin(76.3^\circ)} \approx 8.32 \times 10^{-9} \text{ m}

Since N=1dN = \frac{1}{d}, we have:

N=18.32×10−9≈1.20×108 lines/mN = \frac{1}{8.32 \times 10^{-9}} \approx 1.20 \times 10^8 \text{ lines/m}

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 at higher orders, such as the fifth-order maximum, the peaks may become less distinct due to wider maximums and diffraction effects, which might lead to inaccuracies in measurement.

Step 3

Determine, using Figure 4, V_A for L_R.

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Answer

From Figure 4, we can extrapolate the linear region of the current-voltage characteristic for L_R. The activation voltage VAV_A is indicated at the point where this line intersects the horizontal axis. Based on examination of the graph, VAV_A for L_R is approximately 2.15 V.

Step 4

Deduce a value for the Planck constant.

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Answer

Using the formula: VA=hceρpV_A = \frac{hc}{e \rho_p}, we can rearrange to find hh:

h=VAeρpch = \frac{V_A e \rho_p}{c}

Given:

  • VA=2.00 VV_A = 2.00 \text{ V} for L_G,
  • c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s},
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C},
  • approx hop=650×10−9 m ho_p = 650 \times 10^{-9} \text{ m},

We can substitute these values:

h=(2.00)(1.6×10−19)(650×10−9)(3.00×108)≈6.63×10−34 Jsh = \frac{(2.00)(1.6 \times 10^{-19})(650 \times 10^{-9})}{(3.00 \times 10^8)} \approx 6.63 \times 10^{-34} \text{ Js}

Step 5

Deduce the minimum value of R.

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Answer

Using Ohm's Law: R=VIR = \frac{V}{I} where:

  • V=6.10−VAV = 6.10 - V_A (with VAV_A determined earlier)
  • II is the current, not exceeding 21.0 mA or 0.021 A0.021 \text{ A}.

Substituting:

  • For VA=2.15 VV_A = 2.15 \text{ V}, then: R=6.10−2.150.021≈190.48 OhmsR = \frac{6.10 - 2.15}{0.021} \approx 190.48 \text{ Ohms} Thus, the minimum resistance RR should be at least 191 Ohms.

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