4. A supernova goes off and you can barely detect it with your eyes. Astronomers tell you that supernovae have a luminosity of \(10^{42}\) ergs/s; what is the distance of the supernova?
Assume the supernova emits most of its energy at the peak of the eye's sensitivity and that it explodes isotropically.
From problem 3, we have know the relationship between the luminosity of an object, the flux observed, and the distance between the object and the observer to be: \[ d = (\frac{L_*}{4πF})^{1/2},\] where d is the distance, L is the luminosity, and F is the flux observed.
The problem gives the luminosity of the supernova. In order to solve for the distance, we need to determine the observed flux.
From problem 1, we know that the human eye must receive ~ 10 photons every .1 seconds in order to register the presence of a light. We can call this: \[ N_{\text{photons}} = 100 \text{ photons / s}\] We are also given that the energy of a photon is E = \(h\nu\), for a frequency \(\nu\), and Planck's constant h = 6.6 * \(10^{-27}\) erg*s.
The minimum energy we can receive from the supernova and still register it with our eye is thus: \[ N_{\text{photons}}h\nu,\]with units of \[(\frac{\text{photons}}{\text{s}})(\frac{\text{erg * s *} \frac{1}{\text{s}}}{\text{photon}}) = \frac{\text{erg}}{\text{s}} = \frac{\text{energy}}{{\text{time}}}\]
The quantity we are looking for, flux, needs to have units of energy per time per area, or \[ \frac{\text{energy}}{\text{time * area}}\] Thus, \[ N_{\text{photons}}h\nu = FA\] \[F = \frac{N_{\text{photons}}h\nu}{A}\], where F is the flux observed, \(\nu\) is the frequency of the emitted radiation, h is Planck's constant as above, and A is the area over the the flux is being observed - in this case, the area of a human pupil.
We don't know the frequency of emission from the supernova, but we do know the wavelengths at which the human eye can register light. Using the relationship c = \(\lambda\nu\), where c is the speed of light, \(\lambda\) is the wavelength of the light, and \(\nu\) is the frequency of the light, we can replace \(\nu\) in the equation for \(\frac{c}{\lambda}\): \[F = \frac{N_{\text{photons}}h(\frac{c}{\lambda})}{A}\] \[F = \frac{N_{\text{photons}}hc}{A\lambda}\] We can now plug this expression for the flux into our previous equation for distance: \[ d = (\frac{L_*}{4πF})^{1/2}\] \[ d = (\frac{L_*}{4π\frac{N_{\text{photons}}hc}{A\lambda}})^{1/2}\] \[ d = (\frac{L_*A\lambda}{4π N_{\text{photons}}hc})^{1/2}\]
Now, we plug in the known value for each quantity, approximating the area of the pupil as a circle using a standard pupil radius of 1 mm, and an average wavelength of 500 nm:
\[ d = (\frac{(10^{42} \frac{\text{erg}}{\text{s}})(π (0.1 \text{ cm})^2)(500 \text{ nm} (\frac{1 \text{ cm}}{10^7 \text{ nm}}))}{4π (100 \text{ s}^{-1})(6.67 * 10^{-27} \text{ erg * s})(3 * 10^{10} \frac{\text{ cm}}{\text{ s}})})^{1/2}\]
\[ d = (\frac{(10^{42})(3 (0.1)^2)(5 *10^{-5})}{10 (100)(7 * 10^{-27})(3 * 10^{10})})^{1/2} \text{ cm} = 3 * 10^{24} \text{ cm}\]
\[ d = 3 * 10^{24} \text{ cm} (\frac{1 \text{ pc}}{3 * 10^{18} \text{ cm}})\] \[ d = 9 * 10^5 \text{pc}\] The supernova is \( 9 * 10^5\) pc away.
Fantastic!
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