A light-emitting diode (LED) emits light over a narrow range of wavelengths. These wavelengths are distributed about a peak wavelength $ au_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 $ au$. --- 0 2 . 1 Light from $L_R$ is incident normally on a plane diffraction grating. The fifth-order maximum for light of wavelength $ au_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 $V_A = rac{hc}{e au_p}$ where $h$ = 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 $ au_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 $ au_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 grating formula: $d imes ext{sin}( heta) = m au_p$ where $d$ is the distance between adjacent lines (grating spacing), $heta$ is the diffraction angle, and $m$ is the order number (5 in this case).

First, we calculate $d$ : $d = rac{ au_p}{m imes ext{sin}(76.3^{ ext{o}})}$ Substituting values: Assuming $au_p$ is about 650 nm (near red light), $d = rac{650 imes 10^{-9}}{5 imes ext{sin}(76.3^{ ext{o}})}$ Calculating gives: $N = rac{1}{d} ext{, and thus } N ext{ is approximately } 3.06 imes 10^3 ext{ lines/m}.$

Step 2

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

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Answer

One possible disadvantage of using the fifth-order maximum is that it can be harder to accurately measure the angle due to increased diffraction spread, which can lead to measurement errors. As higher orders can often result in more pronounced overlap of adjacent maxima, leading to difficulties in isolation.

Step 3

Determine, using Figure 4, $V_A$ for $L_R$.

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Answer

To determine the activation voltage $V_A$ for $L_R$ , we extrapolate the linear portion of the current-voltage characteristic curve intersecting the horizontal axis. From Figure 4, we see this intersection occurs at approximately 2.1 V.

Step 4

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

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Answer

From the equation $V_A = \frac{hc}{e \tau_p}$ , rearranging gives: $h = \frac{V_A e \tau_p}{c}$ Substituting the known values, where $V_A$ for $L_G$ is 2.00 V, $e = 1.6 \times 10^{-19} C$ , $c = 3.00 \times 10^8 m/s$ , and using an approximate $au_p$ value of 650 nm (which is $6.50 \times 10^{-7} m$ ) leads to: $h = \frac{(2.00)(1.6 \times 10^{-19})(6.50 \times 10^{-7})}{(3.00 \times 10^8)}$ Calculating this yields a Planck constant value in a range consistent with known values.

Step 5

Deduce the minimum value of $R$.

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Answer

Using Ohm's law and the principle of total voltage in a series circuit, we have: $V = I R + V_A$ Where the total voltage supplied is 6.10 V and $I$ must be less than or equal to 21.0 mA. Substituting gives: $6.10 = (21.0 \times 10^{-3})R + 2.00$ Rearranging to solve for $R$ provides: $R = \frac{6.10 - 2.00}{21.0 \times 10^{-3}}$ Thus gives a minimum resistance of about 195 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

Light from $L_R$ is incident normally on a plane diffraction grating. The fifth-order maximum for light of wavelength $ au_p$ occurs at a diffraction angle of $76.3^{ ext{o}}$. Determine $N$, the number of lines per metre on the grating.

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

Determine, using Figure 4, $V_A$ for $L_R$.

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

Deduce the minimum value of $R$.

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