12. 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 is closed, the circuit is completed and the current can flow through both lamps.

1. The total resistance in the circuit decreases because there are now two paths for the current to take.
2. As a result, the overall current increases, which leads to the following conclusions:
   • The reading on the voltmeter will increase because more voltage is distributed across the combined lower resistance.
   • The voltmeter will measure a higher voltage across the open lamp since it is now drawing current.

When a voltage is applied across an LED, it is forward biased, which allows charge carriers (electrons and holes) to move towards the junction.

1. Electrons from the n-type region move into the p-type region, while holes move into the n-type region, creating recombination.
2. When these electrons fall back to a lower energy state, they release energy in the form of photons, which is the emitted light.
3. The energy of the emitted photon corresponds to the energy band gap, thus determining the color of the light emitted.

To calculate the wavelength of the emitted light, we use the formula:

$E = hf$

where E is the energy of the emitted photon, h is Planck's constant, and f is the frequency. We can rearrange this to find the frequency:

1. From the band gap: $E = 3.03 \times 10^{-19} J$ where $h = 6.63 \times 10^{-34} J s$
2. The frequency is: $f = \frac{E}{h} = \frac{3.03 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.57 \times 10^{14} Hz$
3. Now, to find the wavelength, we can use the speed of light equation: $c = f \cdot \lambda$
4. Rearranging gives us: $\lambda = \frac{c}{f} = \frac{3.00 \times 10^{8}}{4.57 \times 10^{14}} = 6.54 \times 10^{-7} m$

The wavelength calculated is approximately 654 nm, which falls within the visible spectrum. Based on common LED characteristics, this wavelength corresponds to the color red.