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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 $\lambda_p$. Two LEDs L_G an... 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

02.1 Determine N

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Answer

To find the number of lines per metre (N), we use the diffraction formula:

N=1dN = \frac{1}{d}

where dd is the distance between adjacent slits. The angle for the fifth-order maximum is given as 76.3∘^\circ, and the wavelength is read off as λp\lambda_p. Using the formula: d=λN=λpNsin⁡θd = \frac{\lambda}{N} = \frac{\lambda_p}{N \sin \theta}

Thus, we rearrange to find N:

= \frac{1}{\lambda_p \sin(76.3^\circ)} \\ = \frac{1}{\lambda_p \cdot 0.9659} $$ Assuming $\lambda_p$ value corresponds to around 650 nm, substituting gives us: $$ N = \frac{1}{650 \times 10^{-9} \cdot 0.9659} \approx 3.06 \times 10^6 \text{ m}^{-1} $$

Step 2

02.2 Suggest one possible disadvantage

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One disadvantage of using the fifth-order maximum to determine N is that the (5th^{th}) maximum may be wider, making it more difficult to precisely identify the peak compared to lower order maxima. Also, higher-order maxima may be less pronounced or fainter, leading to reduced measurement accuracy.

Step 3

02.3 Determine $V_A$ for L_R

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Answer

Analyzing Figure 4, we identify the point where the linear part of the characteristic for L_R intersects the horizontal axis. By extrapolating this line, we find:

VA for LR≈1.90 VV_A \text{ for } L_R \approx 1.90 \text{ V}

Step 4

02.4 Deduce a value for the Planck constant

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Answer

From the equation given:

VA=hceλpV_A = \frac{hc}{e \lambda_p}

Substituting known values:

  • For L_G, VA=2.00V_A = 2.00 V
  • The wavelength for green light, λp\lambda_p, is approximately 520 nm, and the electron charge, e, is 1.60×10−191.60 \times 10^{-19} C.

Calculating:

h=VAeλpch = \frac{V_A e \lambda_p}{c}

Substituting values into this equation will allow us to calculate the Planck constant.

Step 5

02.5 Deduce the minimum value of R

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Answer

Using Ohm's law, we know:

V=IRV = IR

Given the total voltage across L_R is: 6.10−VA=6.10−2.00=4.10extV6.10 - V_A = 6.10 - 2.00 = 4.10 ext{ V}

The maximum allowable current, I, is 21.0 mA or 0.021 A.

Substituting into Ohm's law, we find:

R=VI=4.100.021≈195.2ΩR = \frac{V}{I} = \frac{4.10}{0.021} \approx 195.2 \Omega

Thus, the minimum value of R must round up to 196 Ω.

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