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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.
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.} $$Step 2
Suggest one possible disadvantage of using the fifth-order maximum to determine N.
Answer
One possible disadvantage of using the fifth-order maximum is that as the order increases, the diffraction pattern becomes less distinct. The maxima may overlap or be less visible due to decreased intensity, making it harder to pinpoint the exact maximum.
Step 3
Step 4
Deduce a value for the Planck constant based on the data given about the LEDs.
Answer
From the relationship given by the formula:
ho_p}$$ 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.
Answer
Using Ohm's law, we can determine the minimum resistance necessary to limit the current through to 21.0 mA. The total voltage from the power supply is 6.10 V.
Setting up the equation based on the required current: Where:
- (the voltage remaining after ),
- .
Thus: R = rac{(6.10 - 2.00)}{21.0 imes 10^{-3}}.
Calculating this will yield the minimum value of resistance .