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 L_G and L_R are adjusted to give the same maximum light intensity. L_G emits green light and L_R emits red light. Figure 3 shows how the light output of the LEDs varies with the wavelength $ ext{λ}$. --- 0 2 . 1 Light from L_R is incident normally on a plane diffraction grating. The fifth-order maximum for light of wavelength $ ext{λ}_p$ occurs at a diffraction angle of 76.3$^ ext{o}$. Determine N, the number of lines per metre on the grating. --- 0 2 . 2 Suggest one possible disadvantage of using the fifth-order maximum to determine N. --- 0 2 . 3 Figure 4 shows part of the current–voltage characteristics for L_R and L_G. When the linear part of the characteristic is extrapolated, the point at which it meets the horizontal axis gives the activation voltage $V_A$ for the LED. $V_A$ for L_G is 2.00 V. Determine, using Figure 4, $V_A$ for L_R. --- 0 2 . 4 It can be shown that $$V_A = \frac{hc}{e\text{λ}_p}$$ where $h$ is the Planck constant. Deduce a value for the Planck constant based on the data given about the LEDs. --- 0 2 . 5 Figure 5 shows a circuit with L_R connected to a resistor of resistance R. The power supply has emf 6.10 V and negligible internal resistance. The current in L_R must not exceed 21.0 mA. Deduce the minimum value of R.

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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

Question 2

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 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

0 2 . 1 Determine N, the number of lines per metre on the grating.

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Answer

To find the number of lines per meter (N) on the grating, we can use the grating equation: $d \sin \theta = n \lambda$ where:

$d$ is the distance between grating lines (grating spacing),
$heta$ is the diffraction angle,
$n$ is the order of the maximum (in this case, n = 5),
$ext{λ}$ is the wavelength.

Given:

$heta = 76.3^ ext{o}$ and
we assume a typical wavelength for red light $ext{λ}_p \approx 650 \text{ nm} = 650 \times 10^{-9} ext{ m}$ .

We can rearrange to find: $d = \frac{n \lambda}{\sin \theta}$

Substituting in the known values gives: $d = \frac{5 \times 650 \times 10^{-9}}{\sin(76.3)} \approx 3.06 \times 10^{-6} \text{ m}$

The number of lines per meter is: $N = \frac{1}{d} = \frac{1}{3.06 \times 10^{-6}} \approx 3.26 \times 10^5 \text{ lines/m}.$

Step 2

0 2 . 2 Suggest one possible disadvantage of using the fifth-order maximum to determine N.

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One possible disadvantage of using the fifth-order maximum to determine N is that higher-order maxima are typically less intense and may be less distinct. As a result, it can be more difficult to accurately identify and measure the position of the fifth-order maximum, which could lead to errors in the calculated value for N.

Step 3

0 2 . 3 Determine, using Figure 4, $V_A$ for L_R.

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To determine the activation voltage $V_A$ for L_R, we need to extrapolate the linear region of the current-voltage characteristic from Figure 4 and read the point where it meets the horizontal axis. From the graph, we find that the activation voltage $V_A$ for L_R is approximately 1.85 V.

Step 4

0 2 . 4 Deduce a value for the Planck constant based on the data given about the LEDs.

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Using the equation: $V_A = \frac{hc}{e\text{λ}_p}$ For L_G, we can rearrange to find: $h = \frac{V_A e \text{λ}_p}{c}$ Where:

$e \approx 1.60 \times 10^{-19} \text{C}$ (charge of an electron),
$c \approx 3.00 \times 10^8 \text{m/s}$ (speed of light), and
$ext{λ}_p \approx 650 \text{nm} = 650 \times 10^{-9} \text{m}$ .

Substituting these values gives: $h \approx \frac{(2.00)(1.60 \times 10^{-19})(650 \times 10^{-9})}{3.00 \times 10^8} \approx 6.63 \times 10^{-34} \text{Js}.$

Step 5

0 2 . 5 Deduce the minimum value of R.

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Using Ohm's Law, the minimum value of the resistor R can be determined as follows: Given the emf from the power supply is 6.10 V and the maximum current is 21.0 mA (or 0.021 A), we can use the equation: $V = IR$ Rearranging gives: $R = \frac{V}{I} = \frac{6.10}{0.021} \approx 290.48 \Omega.$ Thus, the minimum value of R should be at least 290.48 Ω.

A light-emitting diode (LED) emits light over a narrow range of wavelengths - AQA - A-Level Physics - Question 2 - 2021 - Paper 3

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

0 2 . 1 Determine N, the number of lines per metre on the grating.

0 2 . 2 Suggest one possible disadvantage of using the fifth-order maximum to determine N.

0 2 . 3 Determine, using Figure 4, $V_A$ for L_R.

0 2 . 4 Deduce a value for the Planck constant based on the data given about the LEDs.

0 2 . 5 Deduce the minimum value of R.

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