Final answer:
To compare the power of two light sources, we use the energy of a photon and the photon emission rate. The energy of a photon is calculated with the formula E = hc/λ. We cannot solve the problem without the wavelength for S1 but can describe the process using the provided wavelength for S2.
Step-by-step explanation:
The question is asking us to compare the power of two light sources with different photon emission rates and wavelengths. To solve this problem, we need to use the relationship between energy, frequency, and wavelength of light, and recall that power is the rate at which energy is emitted.
The energy E of a photon can be calculated using the formula E = hc/λ, where h is Planck's constant (6.63 × 10-34 J·s), c is the speed of light (3.00 × 108 m/s), and λ is the wavelength of the light. The power emitted by a light source is the product of the energy of each photon and the rate at which photons are emitted (P = E × (number of photons per second)).
Without the wavelength for S1, however, we cannot complete the calculation explicitly. For S2, assuming a wavelength of 5100 Å (or 510 nm), we would calculate the energy per photon and then the power output, but to fully solve the problem, the information about S1 is crucial.