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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 ... 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

Light from L$_R$ is incident normally on a plane diffraction grating. The fifth-order maximum for light of wavelength $ ho_p$ occurs at a diffraction angle of 76.3$^ ext{o}$. 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 diffraction formula:

dsin⁡(θ)=mλd \sin(\theta) = m\lambda

where:

  • d is the distance between grating lines (1/N)
  • heta heta is the diffraction angle (76.3exto^ ext{o})
  • m is the order of the maximum (5)
  • hop ho_p is the wavelength corresponding to the maximum

From the graph (Figure 3), we read off hop≈650 nm=650×10−9 m ho_p \approx 650 \text{ nm} = 650 \times 10^{-9} \text{ m}. Now, rearranging the equation:

N=mλdsin⁡(θ)N = \frac{m \lambda}{d \sin(\theta)}

Substituting known values: N=5×650×10−9sin⁡(76.3exto)N = \frac{5 \times 650 \times 10^{-9}}{\sin(76.3^ ext{o})}

Calculating gives: N≈3.06×103 lines/mN \approx 3.06 \times 10^3 \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 the fifth-order maximum may be weaker or less defined compared to lower-order maxima. This could lead to less accurate measurements due to potential interference or greater uncertainty in determining the exact angle.

Step 3

Determine, using Figure 4, V$_A$ for L$_R$.

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Answer

From Figure 4, we observe the current-voltage (I-V) characteristic curve of LR_R. To find the activation voltage VA_A, we look for the point where the linear part of the curve intersects the x-axis (V-axis). By extrapolating the linear section of the graph, we find that VA≈1.93 V_A \approx 1.93 \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 relationship: VA=hceρpV_A = \frac{hc}{e\rho_p} We rearrange to find h: h=VAeρpch = \frac{V_A e \rho_p}{c} Substituting:

  • VA_A for LG_G: 2.00 V
  • e: 1.60×10−19 C1.60 \times 10^{-19} \text{ C}
  • hop ho_p: 650 nm = 650×10−9 m650 \times 10^{-9} \text{ m}
  • c: 3.00×108 m/s3.00 \times 10^8 \text{ m/s}

Calculating: h=(2.00)(1.60×10−19)(650×10−9)3.00×108h = \frac{(2.00)(1.60 \times 10^{-19})(650 \times 10^{-9})}{3.00 \times 10^8} h≈6.63×10−34 Jsh \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, the relationship between voltage (V), current (I), and resistance (R) is given by: R=VIR = \frac{V}{I} Given:

  • Voltage from power supply = 6.10 V
  • Maximum current in LR_R must not exceed 21.0 mA = 0.021 A

Calculating minimum R: Rmin=6.100.021≈290.48ΩR_{min} = \frac{6.10}{0.021} \approx 290.48 \Omega Thus, the minimum value of R is approximately 290.48 Ω.

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