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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 $ heta_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, we can use the diffraction grating equation: dsin⁡θ=nλd \sin \theta = n \lambda where:

  • dd is the distance between adjacent slits (lines),
  • θ\theta is the diffraction angle,
  • nn is the order of the maximum (in this case, 5),
  • λ\lambda is the wavelength of the light.

First, rearranging gives: d=nλsin⁡θd = \frac{n \lambda}{\sin \theta}

We read from the graph the peak wavelength, λp\lambda_p, can be approximated as 650\nm=650×10−9m650 \nm = 650 \times 10^{-9} m.

Substituting in:

= \frac{3.25 \times 10^{-6} m}{0.9659} \\ \approx 3.36 \times 10^{-6} m$$ Now, since $N = \frac{1}{d}$, we can compute: $$N = \frac{1}{3.36 \times 10^{-6}} \approx 297000 \ m^{-1} = 2.97 \times 10^5 \ m^{-1}$$

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 the peaks may be less distinct and harder to measure accurately compared to lower orders. As the order increases, the intensity of the maximum tends to decrease, leading to potential issues in obtaining precise measurements.

Step 3

Determine, using Figure 4, V_A for L_R.

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Answer

From Figure 4, to find the activation voltage VAV_A for LRL_R, we need to extrapolate the linear region of the current-voltage characteristic of LRL_R to where it intersects the horizontal axis. Based on observation, this gives approximately VA≈1.85VV_A \approx 1.85 V.

Step 4

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

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Answer

Given the equation VA=hceθpV_A = \frac{hc}{e \theta_p} We can rearrange this to find Planck's constant (hh): h=VAeθpch = \frac{V_A e \theta_p}{c} We know:

  • VAV_A for LG=2.00VL_G = 2.00 V
  • Elementary charge e≈1.6×10−19Ce \approx 1.6 \times 10^{-19} C
  • Speed of light c=3.0×108m/sc = 3.0 \times 10^8 m/s
  • Peak wavelength hetap≈550nm=550×10−9m heta_p \approx 550 nm = 550 \times 10^{-9} m

Substituting these values in:

\approx 5.83 \times 10^{-34} J \, s$$

Step 5

Deduce the minimum value of R.

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Answer

From the circuit presented in Figure 5, we know:

  • The total voltage supplied, Vs=6.10VV_s = 6.10 V.
  • The maximum current through LRL_R must not exceed 21.0mA=0.021A21.0 mA = 0.021 A.
  • To find the minimum resistance RR, we can use Ohm's law: V=IRV = IR Rearranging gives: R=Vs−VAIR = \frac{V_s - V_A}{I} Substituting in:
= \frac{4.10 V}{0.021 A} \approx 195.24 \Omega$$ Thus, the minimum value of resistance, rounded appropriately, is $195 \Omega$.

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