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?

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Answer 1 · 2018-03-21T01:31:33

#6.83xx10^(-11)"J"/"hr"#

The energy of a photon is:

#E_\lambda = (hc)/\lambda#

This means the energy per photon is:

#E_\lambda=((6.626 xx 10^-34)(3xx10^8))/(3.35xx10^(-3))=5.93xx10^(-23) "J"/"photon"#

The detector is capturing #3.2 xx 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") )#
#=1.152xx10^(12) "photons"/"hr"#

Total energy in 1 hour detected is:

#E =(1.152xx10^(12) "photons"/"hr") E_\lambda=(1.152xx10^(12) cancel"photons"/"hr") (5.93xx10^(-23) "J"/cancel"photon")#
#=6.83xx10^(-11)"J"/"hr"#