An object is emitting radiation at 1520 nm. A detector is capturing 7 x10^7 photons/sec at this wavelength. h = 6.63x10-34 J-s. c = 2.998 x 108 m/s. Total energy of the photons detected in 1 sec?

Answer 1

Wavelength = λ = 1520 nm = 1520x10^-9 m = 1.520x10^-6 m

frequency = ν = ?

c = speed of light = 2.998x10⁸ m/s

E = energy = ?

First, we'll find the ν

c = λν

ν = c/λ = 2.998x10⁸ m/s / 1.520x10^-6 m = 1.97x10¹⁴ s^-1

Now, we can solve for E (energy):

E = hν = (6.63x10^-34 Js)(1.97x10¹⁴ s^-1 ) = 1.31x10^-19 J ... this is the energy per photon

Total energy detected in 1 second = 7x10⁷ photons/s x 1.31x10^-19 J/photon = 9.17x10^-12 J/sec

Answer 2

The energy of one photon is given by E = hc/λ where h is Planck's constant, c the speed of light, and λ is the wavelength.

E = 6.63 x 10^-34 J-s x 2.998 x 10⁸ m/sec / 1250 x 10^-9 m. = 1.59^-19 J/photon

As there are 7 x 10⁷ photons detected per second, the total photon energy is

7 x 10⁷ photons s^-1 x 1.59 x 10^-19 J/photon = 1.11 x 10^-11 J/s