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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 LG 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

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) on the grating, we use the diffraction grating formula: dsin⁡θ=nλd \sin \theta = n\text{λ} where d is the distance between the lines, heta heta is the angle of diffraction, n is the order of the maximum, and extλ ext{λ} is the wavelength. Given the angle heta=76.3∘ heta = 76.3^{\circ} and n = 5 (fifth-order), we can first find extλp ext{λ}_p from Figure 3, taking it to be approximately 650 nm (red light).

Substituting: d=λsin⁡θ=650×10−9msin⁡(76.3∘)d = \frac{\text{λ}}{\sin \theta} = \frac{650 \times 10^{-9} \text{m}}{\sin(76.3^{\circ})} We find: d≈650×10−9m/0.9659≈672×10−9md \approx 650 \times 10^{-9} \text{m} / 0.9659 \approx 672 \times 10^{-9} \text{m} N is then calculated as: N=1d≈1672×10−9m≈1.48×106lines/mN = \frac{1}{d} \approx \frac{1}{672 \times 10^{-9} \text{m}} \approx 1.48 \times 10^{6} \text{lines/m}

Step 2

Suggest one possible disadvantage of using the fifth-order maximum to determine N.

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Answer

One disadvantage of using the fifth-order maximum is that it may not be well-defined due to wider peaks, which could lead to inaccuracies in measuring the angle and thus determining N.

Step 3

Determine, using Figure 4, VA for LR.

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Answer

To find the activation voltage extVA ext{V}_A for LR, we need to extrapolate the linear part of the current-voltage characteristic until it meets the horizontal axis. The value at the intersection point, according to Figure 4, is approximately 2.12 V.

Step 4

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

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Using the relationship VA=hceλp\text{V}_A = \frac{hc}{e\text{λ}_p}, we can rearrange this to find Planck's constant (h): h=VAeλpch = \frac{\text{V}_A e \text{λ}_p}{c} Substituting in values:

  • extVAextforLR≈2.12extV ext{V}_A ext{ for LR} \approx 2.12 ext{ V},
  • e≈1.60×10−19extCe \approx 1.60 \times 10^{-19} ext{ C},
  • extλp≈650×10−9extm ext{λ}_p \approx 650 \times 10^{-9} ext{ m},
  • c≈3.00×108extm/sc \approx 3.00 \times 10^8 ext{ m/s}, we find: h≈(2.12)(1.60×10−19)(650×10−9)3.00×108≈1.05×10−34extJsh \approx \frac{(2.12)(1.60 \times 10^{-19})(650 \times 10^{-9})}{3.00 \times 10^8} \approx 1.05 \times 10^{-34} ext{ J s}

Step 5

Deduce the minimum value of R.

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Answer

To find the minimum value of R, we use Ohm's Law: V=IRV = IR Given that the power supply voltage is 6.10 V and the maximum current in LR is 21.0 mA, we can find: Using I=21.0×10−3 AI = 21.0 \times 10^{-3} \text{ A}, we rearrange for R: R=VI=6.10extV21.0×10−3extA≈290.48 ΩR = \frac{V}{I} = \frac{6.10 ext{ V}}{21.0 \times 10^{-3} ext{ A}} \approx 290.48 \text{ Ω} Thus, the minimum value of R is approximately 290 Ω.

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