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$. Figure 3: *graph of light intensity vs wavelength* --- Light from L$_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$_G$. Figure 4: *graph of current vs voltage* 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$_R$. --- It can be shown that $$V_A = \frac{hc}{e\rho_p}$$ 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. Figure 5: *schematic diagram of circuit* 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 $ 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:

$d \sin(\theta) = m\lambda$

where:

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

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

$N = \frac{m \lambda}{d \sin(\theta)}$

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

Calculating gives: $N \approx 3.06 \times 10^3 \text{ lines/m}$

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 L $_R$ . To find the activation voltage V $_A$ , 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 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: $V_A = \frac{hc}{e\rho_p}$ We rearrange to find h: $h = \frac{V_A e \rho_p}{c}$ Substituting:

V $_A$ for L $_G$ : 2.00 V
e: $1.60 \times 10^{-19} \text{ C}$
$ho_p$ : 650 nm = $650 \times 10^{-9} \text{ m}$
c: $3.00 \times 10^8 \text{ m/s}$

Calculating: $h = \frac{(2.00)(1.60 \times 10^{-19})(650 \times 10^{-9})}{3.00 \times 10^8}$ $h \approx 6.63 \times 10^{-34} \text{ Js}$

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: $R = \frac{V}{I}$ Given:

Voltage from power supply = 6.10 V
Maximum current in L $_R$ must not exceed 21.0 mA = 0.021 A

Calculating minimum R: $R_{min} = \frac{6.10}{0.021} \approx 290.48 \Omega$ Thus, the minimum value of R is approximately 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

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.

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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