Question

The ratio of the power of a light source \( S_1 \) to that of the light source \( S_2 \) is 2. \( S_1 \) is emitting \( 2 \times 10^{15} \) photons per second at 600 nm. If the wavelength of the source \( S_2 \) is 300 nm, then the number of photons per second emitted by \( S_2 \) is \_\_\_\_\_ \( \times 10^{14} \).

Hint:

The number of photons emitted is inversely proportional to the energy of each photon. For shorter wavelengths, the energy per photon is higher, so fewer photons are emitted for the same power.

Answer 1

The power of a light source is related to the energy emitted by the source per second, which is directly proportional to the number of photons emitted per second. The energy of a photon is given by: \[ E = \frac{hc}{\lambda}, \] where: - \( h \) is Planck's constant, - \( c \) is the speed of light, - \( \lambda \) is the wavelength. The power ratio is given by: \[ \frac{P_1}{P_2} = \frac{N_1 E_1}{N_2 E_2}, \] where \( N_1 \) and \( N_2 \) are the number of photons emitted by \( S_1 \) and \( S_2 \), respectively, and \( E_1 \) and \( E_2 \) are the energies of the photons emitted by \( S_1 \) and \( S_2 \), respectively. Given that the power ratio is 2, and the wavelength of \( S_1 \) is 600 nm and \( S_2 \) is 300 nm, we know that the energy of a photon is inversely proportional to its wavelength. Therefore: \[ E_2 = 2 E_1. \] Now, using the relation for the number of photons and the given values: \[ \frac{N_1}{N_2} = \frac{2 \times 10^{15}}{2} \times \frac{600}{300} = 8 \times 10^{14}. \] Final Answer: \( 8 \times 10^{14} \).