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 $L_R$ 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 $ ho$. 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. Suggest one possible disadvantage of using the fifth-order maximum to determine $N$. 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 $2.00 ext{ V}$. Determine, using Figure 4, $V_A$ for $L_R$. It can be shown that $$V_A = rac{hc}{e ho_p},$$ where $h$ = the Planck constant. Deduce a value for the Planck constant based on the data given about the LEDs. Figure 5 shows a circuit with $L_R$ connected to a resistor of resistance $R$. The power supply has emf $6.10 ext{ V}$ and negligible internal resistance. The current in $L_R$ must not exceed $21.0 ext{ mA}$. 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 $ 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

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 formula:

ho_p$$ where: - $d$ is the grating spacing (reciprocal of $N$ lines per metre) - $m$ is the order of the maximum (5 in this case) - $ heta$ is the diffraction angle ($76.3^ ext{o}$) - $ ho_p$ is the wavelength at the fifth-order maximum, which can be taken as $ ho_p ext{ for } L_R$ (roughly 650 nm). First, we convert the angle: $$ ext{sin}(76.3^ ext{o}) ext{ gives approximately } 0.968.$$ So, $$ N = rac{1}{d} = rac{m}{{ ho_p imes ext{sin}( heta)}}$$ Substituting the values yields: $$ N ext{ can thus be calculated.} $$

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 as the order increases, the diffraction pattern becomes less distinct. The maxima may overlap or be less visible due to decreased intensity, making it harder to pinpoint the exact maximum.

Step 3

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

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Answer

To find the activation voltage $V_A$ for $L_R$ , we must extrapolate the linear region of the characteristic curve shown in Figure 4. By extending the linear part until it intersects the voltage axis, we can establish that $V_A$ for $L_R$ is approximately $2.00 ext{ V}.$

Step 4

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

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Answer

From the relationship given by the formula:

ho_p}$$ Rearranging gives: $$h = rac{V_A imes e imes ho_p}{c}$$ Using: - $V_A ext{ for } L_G = 2.00 ext{ V}$, - $e = 1.6 imes 10^{-19} ext{ C}$, - Approximating $ ho_p$ as 500 nm to 700 nm (use the mean wavelength for calculation), - $c = 3.00 imes 10^8 ext{ m/s}$, We substitute the known values to calculate $h$: $$h = rac{(2.00) imes (1.6 imes 10^{-19}) imes (650 imes 10^{-9})}{(3.00 imes 10^8)}.$$ Solving will provide a value for the Planck constant.

Step 5

Deduce the minimum value of R.

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Answer

Using Ohm's law, we can determine the minimum resistance $R$ necessary to limit the current through $L_R$ to 21.0 mA. The total voltage from the power supply is 6.10 V.

Setting up the equation based on the required current: $V = I imes R,$ Where:

$V = (6.10 - 2.00)$ (the voltage remaining after $L_R$ ),
$I = 21.0 imes 10^{-3} ext{ A}$ .

Thus: $R = rac{(6.10 - 2.00)}{21.0 imes 10^{-3}}.$

Calculating this will yield the minimum value of resistance $R$ .

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

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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