Question

A stellar object is emitting radiation at 1970 nm. A detector is capturing 9 107 photons per second at this wavelength. h = 6.63 x 10-34 J-s. c = 2.998 x 108 m/s. (a) What is the total energy of the photons detected in one second? J/s (b) What is the total energy of the photons detected in one hour? J/hr

Answer 1

Determine the energy of ONE PHOTON. Then multiply that amount by 9101.
Photon Energy = hν = hc/λ

h = Planck's constant (6.636 x10^(-34) J*s
ν = frequency
c = speed of light in vacuo (2.9979 x10^(8)
λ = wavelength = 1.97 x 10^(-6) m

Single Photon
E = hc/λ = (6.636 x10^(-34) J*s) * (2.9979 x10^(8) m/s) / 1.97 x 10^(-6) = 1.00985x10^(-19) J

Total Photon Energy = (1.00985x10^(-19) J/photon) * 9107 photons = 9.196 x10^(-16) J

You can express that in eV, too: 9.196 x10^(-16) / 1.602x10^(-19) = 5.74 x 10³ eV per second & 3.44 x 10⁵ ev per hour