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 equates to the energy it emits per second, which is directly proportional to the photon emission rate. A photon's energy is determined by: \[ E = \frac{hc}{\lambda}, \] where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. The ratio of powers is expressed as: \[ \frac{P_1}{P_2} = \frac{N_1 E_1}{N_2 E_2}, \] where \( N_1 \) and \( N_2 \) represent the photon emission counts from sources \( S_1 \) and \( S_2 \), respectively, and \( E_1 \) and \( E_2 \) are the corresponding photon energies. Given a power ratio of 2, and wavelengths of \( S_1 \) at 600 nm and \( S_2 \) at 300 nm, photon energy is inversely proportional to wavelength. Consequently: \[ E_2 = 2 E_1. \] Utilizing the relationship for photon count and the provided values: \[ \frac{N_1}{N_2} = \frac{2 \times 10^{15}}{2} \times \frac{600}{300} = 8 \times 10^{14}. \] Final Answer: \( 8 \times 10^{14} \).