Question

A sodium vapor lamp has a power output of 300 W. If 590 nm is the average wavelength of the photons, about how many photons are emitted per second?

Answer 1

Power is measured in Watts (W), which is the same as Joules per second (J/s), i.e. the rate of energy loss/emission by the object.

The energy of a photon of light is given by

E = hf

where h is Planck's constant and f is the frequency of the light. But we can also write this as

E = hc / ?

because frequency, f, is equal to the speed of light, c, divided by the wavelength, ?. Putting in the numbers,

E = (6.63x10^-34)(3x10^8) / (590x10^-9) = 3.371x10^-19 J.

Now, remember that I said that power is the same as the rate of energy loss/emission. That means that power is the amount of energy lost/emitted over a certain period of time, or, as an equation,

P = dE/dt = E/t in this case,

since we are talking about absolute values of E and t, not the difference between two values. Putting in our values for P and E then gives

300 = 3.371x10^-19 / t.

Now we can rearrange and simply to get

t = 3.371x10^-19 / 300 = 11.236x10^-22 s.

Since we want to know the number of photons emitted per second (let's call it "n"), then all we need to do is to write

n = 1/t = 1 /11.236x10^-22 =0.8899x10^22 photons per second (photons/s).