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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 $ ext{... 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 determine the number of lines per metre (N) on the grating, we first use the diffraction grating formula: dsin⁡θ=mλd \sin \theta = m \lambda where:

  • dd is the distance between the slits (grating spacing),
  • heta heta is the diffraction angle,
  • mm is the order of the maximum (5 for the fifth-order maximum),
  • \lambda is the wavelength of the light.

From the problem statement, we will read off the corresponding extλp ext{λ}_p for red light. For this example, let’s assume extλp ext{λ}_p = 650 nm:

Convert this to metres: extλp=650×10−9 m. ext{λ}_p = 650 \times 10^{-9} \, m.

Substituting into the equation yields: N=1d=1λp/sin⁡(76.3∘)N = \frac{1}{d} = \frac{1}{\text{λ}_p / \sin(76.3^\circ)}

After calculations: N=3.06×103 m−1.N = 3.06 \times 10^3 \, 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 of using the fifth-order maximum to determine N is that in higher-order maxima, the intensity of light can be significantly reduced. This makes it more difficult to accurately measure the diffraction angle, leading to potential inaccuracies in the calculation of N.

Step 3

Determine, using Figure 4, $V_A$ for $ ext{L}_R$.

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Answer

To determine VAV_A for extLR ext{L}_R, we find the point at which the linear part of the characteristic intersects the horizontal axis in Figure 4.

By extrapolating this linear section, we find that the activation voltage VAV_A for extLR ext{L}_R is approximately 1.94 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=hceλp,V_A = \frac{h c}{e \lambda_p}, we can rearrange to find hh: h=VAeλpc.h = \frac{V_A e \lambda_p}{c}. Given that VAV_A for extLG ext{L}_G is 2.00 V and assuming e=1.60×10−19 Ce = 1.60 \times 10^{-19} \, C and c=3.00×108 m/sc = 3.00 \times 10^8 \, m/s, substituting extλp=650×10−9 m ext{λ}_p = 650 \times 10^{-9} \, m:

h=(2.00)(1.60×10−19)(650×10−9)3.00×108.h = \frac{(2.00)(1.60 \times 10^{-19})(650 \times 10^{-9})}{3.00 \times 10^8}.

Calculating this gives: h≈6.63×10−34 J s.h \approx 6.63 \times 10^{-34} \, J \, s.

Step 5

Deduce the minimum value of R.

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Answer

Given that the power supply has an emf of 6.10 V and the current in extLR ext{L}_R must not exceed 21.0 mA (or 0.021 A), we can use Ohm's Law to determine the minimum value of the resistor R:

From Ohm's Law: V=IR⇒R=VI.V = I R \Rightarrow R = \frac{V}{I}. Substituting in our known values: R=6.100.021≈290.48 Ω.R = \frac{6.10}{0.021} \approx 290.48 \, \Omega. Rounding up, the minimum value of R must be at least 291 Ω.

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