A light-emitting diode (LED) emits light over a narrow range of wavelengths - AQA - A-Level Physics - Question 2 - 2021 - Paper 3

One possible disadvantage is that the fifth-order maximum may be less pronounced or fainter, making it harder to measure accurately compared to lower order maxima, which could result in less reliable data.

From the linear region of the ( L_R ) characteristic, we can extrapolate to find the activation voltage ( V_A ).

It appears from the graph that ( V_A \approx 1.85 V ) for the red LED ( L_R ).

Using the formula for the activation voltage:

$V_A = \frac{hc}{e\lambda_p},$

Where:

• ( V_A ) is known for ( L_G ) (2.00 V).
• Using a wavelength of approximately 550 nm for green light: ( \lambda_p = 550 \times 10^{-9} ) m.
• Using the charge of an electron ( e \approx 1.6 \times 10^{-19} ) C.

Rearranging to find ( h ):

$h = \frac{V_A e \lambda_p}{c} = \frac{2.00 \times (1.6 \times 10^{-19}) \times (550 \times 10^{-9})}{3.00 \times 10^8}$

Calculating gives:

$h \approx 6.63 \times 10^{-34} \text{ J s}.$

Using Ohm's law and the power supply specifications:

The voltage across the resistor is: $V_R = V_{source} - V_A = 6.10 - 2.00 = 4.10 V.$

The maximum current through ( L_R ) is 21.0 mA:

$I_{max} = 0.021 A.$

Using Ohm's law to find resistance: $R = \frac{V_R}{I_{max}} = \frac{4.10}{0.021} \approx 195.24 \Omega.$

Thus, the minimum value of ( R ) should be rounded up to 196 Ω.