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

2.406*10^11 eV

Explanation:

The formula for finding the energy of a photon is;

E_gamma=(hc)/lamda

Where h is Planck's constant, c is the speed of light in a vacuum, and lamda is the wavelength of the photon. The energy of photons is typically expressed in terms of electron volts, eV, where 1 "eV" = 1.602*10^-19 "J" is the amount of energy required to accelerate an electron through one volt.

Planck's constant in terms of electron volts is 4.136xx10^-15 "eV s", and the speed of light in a vacuum is 2.998 xx 10^8 "m/s", so the energy for a single photon is;

E_gamma = ((4.136 xx 10^-15"eV s")(2.998*10^8 "m/s"))/(1.670 * 10^-6 "m")

E_gamma = .7425 "eV"

We are given the rate, nu = 9xx10^7 "/s", at which photons are received, so the rate at which energy is received can be stated as;

p = E_gamma*nu

p=(.7425 "eV")(9*10^7"/s")

p=6.682xx10^7 "eV/s"

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

E_T = p* t

E_T = (6.682 xx 10^7 "eV/s")(3600 "s")

2.406*10^11 eV

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