# A stellar object is emitting radiation at 3.35 mm? If the detector is capturing 3.2×10^8 photons per second at this wavelength, what is the total energy of the photons detected in one hour?

Mar 21, 2018

$6.83 \times {10}^{- 11} \text{J"/"hr}$

#### Explanation:

The energy of a photon is:

${E}_{\setminus} \lambda = \frac{h c}{\setminus} \lambda$

This means the energy per photon is:

${E}_{\setminus} \lambda = \frac{\left(6.626 \times {10}^{-} 34\right) \left(3 \times {10}^{8}\right)}{3.35 \times {10}^{- 3}} = 5.93 \times {10}^{- 23} \text{J"/"photon}$

The detector is capturing $3.2 \times {10}^{8}$ photons per second. This corresponds to:

3.2xx10^8 ("photons" / cancel"s")((60 cancel" s") / (1 cancel" min")) ((60 cancel" min") / (1 " hr") )
$= 1.152 \times {10}^{12} \text{photons"/"hr}$

Total energy in 1 hour detected is:

$E = \left(1.152 \times {10}^{12} \text{photons"/"hr") E_\lambda=(1.152xx10^(12) cancel"photons"/"hr") (5.93xx10^(-23) "J"/cancel"photon}\right)$
$= 6.83 \times {10}^{- 11} \text{J"/"hr}$