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?

Question

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?

Owen Atkinson · Accepted Answer

#2.406*10^11 eV#

The equation used to calculate a photon's energy 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#

In contrast, this represents a billionth of the energy released every second by a 100 watt lightbulb.

Hailey Alvarez · Answer

undefined To find the total energy of the photons detected in one hour, first calculate the energy of one photon using the formula E = hc/λ, where E is energy, h is Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength in meters. Then, multiply the energy of one photon by the number of photons detected per second, and finally multiply by the number of seconds in one hour (3600 seconds).

$$ E = \frac{hc}{\lambda} $$

$$ E = \frac{(6.626 \times 10^{-34} \, \text{J*s})(3.00 \times 10^8 \, \text{m/s})}{1670 \times 10^{-9} \, \text{m}} $$

$$ E = \frac{(1.988 \times 10^{-25} \, \text{J*m})}{1670 \times 10^{-9} \, \text{m}} $$

$$ E = 1.19 \times 10^{-19} \, \text{J} $$

$$ \text{Total energy of photons detected in one second} = (1.19 \times 10^{-19} \, \text{J}) \times (9 \times 10^7 \, \text{photons/second}) $$

$$ \text{Total energy of photons detected in one hour} = \text{Total energy of photons detected in one second} \times (3600 \, \text{seconds/hour}) $$

$$ \text{Total energy of photons detected in one hour} = (1.19 \times 10^{-19} \, \text{J}) \times (9 \times 10^7 \, \text{photons/second}) \times (3600 \, \text{seconds/hour}) $$

$$ \text{Total energy of photons detected in one hour} = 3.83 \times 10^{-11} \, \text{J} $$