A stellar object is emitting radiation at 1670 nm. If the detector is capturing 9 * 10^7 photons per second at this wavelength, what is the total energy of the photons detected in one hour?

1 Answer
Dec 19, 2015

2.4061011eV

Explanation:

The formula for finding the energy of a photon is;

Eγ=hcλ

Where h is Planck's constant, c is the speed of light in a vacuum, and λ is the wavelength of the photon. The energy of photons is typically expressed in terms of electron volts, eV, where 1eV=1.6021019J is the amount of energy required to accelerate an electron through one volt.

Planck's constant in terms of electron volts is 4.136×1015eV s, and the speed of light in a vacuum is 2.998×108m/s, so the energy for a single photon is;

Eγ=(4.136×1015eV s)(2.998108m/s)1.670106m

Eγ=.7425eV

We are given the rate, ν=9×107/s, at which photons are received, so the rate at which energy is received can be stated as;

p=Eγν

p=(.7425eV)(9107/s)

p=6.682×107eV/s

Now we can calculate the total energy received over an hour. There are 3600 seconds in an hour.

ET=pt

ET=(6.682×107eV/s)(3600s)

2.4061011eV

For comparison, this is about 1 billionth of the energy given off by a 100 watt light bulb each second.