# 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?

Dec 19, 2015

$2.406 \cdot {10}^{11} e V$

#### Explanation:

The formula for finding the energy of a photon is;

${E}_{\gamma} = \frac{h c}{l} a m \mathrm{da}$

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

Planck's constant in terms of electron volts is $4.136 \times {10}^{-} 15 \text{eV s}$, and the speed of light in a vacuum is $2.998 \times {10}^{8} \text{m/s}$, 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")

${E}_{\gamma} = .7425 \text{eV}$

We are given the rate, $\nu = 9 \times {10}^{7} \text{/s}$, at which photons are received, so the rate at which energy is received can be stated as;

$p = {E}_{\gamma} \cdot \nu$

$p = \left(.7425 \text{eV")(9*10^7"/s}\right)$

$p = 6.682 \times {10}^{7} \text{eV/s}$

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

${E}_{T} = p \cdot t$

${E}_{T} = \left(6.682 \times {10}^{7} \text{eV/s")(3600 "s}\right)$

$2.406 \cdot {10}^{11} e V$

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