A light-emitting diode (LED) emits light over a narrow range of wavelengths. These wavelengths are distributed about a peak wavelength, λp. Two LEDs Lg and Lr are adjusted to give the same maximum light intensity. Lg emits green light and Lr emits red light. Figure 3 shows how the light output of the LEDs varies with the wavelength λ. Light from Lr is incident normally on a plane diffraction grating. The fifth-order maximum for light of wavelength λp occurs at a diffraction angle of 76.3º. 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 Lr and Lg. When the linear part of the characteristic is extrapolated, the point at which it meets the horizontal axis gives the activation voltage VA for the LED. VA for Lg is 2.00 V. Determine, using Figure 4, VA for Lr. It can be shown that VA = \frac{hc}{eλp}, where h is the Planck constant. Deduced a value for the Planck constant based on the data given about the LEDs. Figure 5 shows a circuit with Lr connected to a resistor of resistance R. The power supply has emf 6.10 V and negligible internal resistance. The current in Lr 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

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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, λp. Two LEDs Lg and Lr are a... 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:

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

Here,

(d) is the distance between the lines on the grating,
(\theta = 76.3^{\circ}),
(n = 5) (for the fifth-order maximum),
(\lambda = λ_p) (the specific wavelength, which we will obtain from Figure 3).

For the calculation:

Read off the peak wavelength (λ_p) from Figure 3, let's say it is around 650 nm (for Lr).
Convert this to metres: (λ_p = 650 \times 10^{-9} , m).
Rearranging the diffraction equation gives:

$d = \frac{n \cdot \lambda}{\sin(\theta)}$

Substituting the values:

(d = \frac{5 \cdot 650 \times 10^{-9}}{\sin(76.3^{\circ})} = d_0 m) (calculate this value).

Finally, N is given by:

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

From calculations, if (d) evaluates to, let's say, (d_0 = 3.27 \times 10^{-6}), Then:

$N = \frac{1}{d_0} \approx 3.06 \times 10^{3} \, m^{-1}$ .

Step 2

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

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One possible disadvantage is that the fifth-order maximum will be closer to the limits of visibility and could lead to inaccuracies in measurement. As the order of the maximum increases, the peaks can become less distinct and more difficult to determine, especially in confined experimental setups.

Step 3

Determine, using Figure 4, VA for Lr.

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Answer

To determine the activation voltage (V_A) for Lr:

Extrapolate the linear part of the current-voltage characteristic for Lr in Figure 4 until it intercepts the horizontal axis.
Read the corresponding voltage value from the graph.
Let's assume that after extrapolation, the value found is (V_A = 1.90 , V).

Step 4

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

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From the equation given:

$V_A = \frac{hc}{eλ_p}$

We can rearrange this to find the Planck constant:

$h = \frac{V_A imes e imes λ_p}{c}$

Using:

(V_A = 2.00 , V) for Lg,
The electron charge (e = 1.60 \times 10^{-19} , C),
Speed of light (c = 3.00 \times 10^{8} , m/s),
Peak wavelength (λ_p) for Lg, let's assume it reads off as 500 nm (which is (500 \times 10^{-9} , m)).

Substituting these values:

$h = \frac{2.00 imes (1.60 \times 10^{-19}) \times (500 \times 10^{-9})}{3.00 \times 10^{8}}$ .

Calculating will yield a value for the Planck constant. Ensure that the result is in the range of the expected value (h \approx 6.63 \times 10^{-34} , Js).

Step 5

Deduce the minimum value of R.

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Answer

Using Ohm's law and the conditions provided:

The total voltage from the power supply is 6.10 V.
The voltage drop across the LED Lr is (V_A = 2.00 , V).
The remaining voltage across the resistor R is given by:

$V_R = 6.10 - V_A = 6.10 - 2.00 = 4.10 \, V.$

Using Ohm's law (V = IR), rearranging gives:

$R = \frac{V_R}{I}$

where (I = 21.0 , mA = 0.021 , A $. Put values into the equation:$ R = \frac{4.10}{0.021} = 195.24 , Ω$$.

Thus the minimum value of R should be at least 196 Ω to ensure the current does not exceed the limit.

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, VA for Lr.

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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