A technician sets up a circuit as shown, using a car battery and two identical lamps - Scottish Highers Physics - Question 12 - 2016

When switch S closes, the reading on the voltmeter decreases. This is due to the fact that closing the switch allows current to flow through the circuit, increasing the total current and thus decreasing the voltage across the lamps due to the increased power demand. Consequently, the voltage drops across the fixed resistances in the circuit which is reflected on the voltmeter.

When a voltage is applied across the LED, it becomes forward biased, meaning the charge carriers (electrons from the n-type and holes from the p-type) move towards the junction. Electrons from the n-type region drop into the conduction band, and as they move from the conduction band to the valence band of the p-type region, they recombine with holes. During this recombination process, photons of light are emitted, corresponding to the energy difference between the two bands.

To calculate the wavelength, we use the formula that relates energy and wavelength:

E = rac{hc}{ ext{wavelength}}

Where:

• $E = 3.03 \times 10^{-19} \text{ J}$
• $h = 6.63 \times 10^{-34} \text{ J s}$ (Planck's constant)
• $c = 3 \times 10^{8} \text{ m/s}$ (speed of light)

Rearranging the equation for wavelength gives:

$\text{wavelength} = \frac{hc}{E}$

Substituting the values:

$\text{wavelength} = \frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{3.03 \times 10^{-19}}$

Calculating gives:

$\text{wavelength} = 6.56 \times 10^{-7} \text{ m}$.

The wavelength calculated is approximately 650 nm, which falls within the red spectrum of visible light. Therefore, the colour of the light emitted by the LED is red.