This feature arises from hydrogen gas in the accretion disk. The photons radiated during the accretion process are constantly ionizing nearby hydrogen atoms. So there are many free protons and electrons in the disk. When one of these protons comes close enough to an electron, they recombine into a new hydrogen atom, and the electron will lose energy until it reaches the lowest allowed energy state (called the ground state), labeled n = 1.
On its way to the ground state, the electron passes through other allowed energy states (called excited states). Technically speaking, atoms have an infinite number of allowed energy states, but electrons spend most of their time occupying those of lowest energies.
Because the difference in energy between. e.g., the n = 2 and n = 1 states are always the same, the electron always loses the same amount of energy when it passes between them. Thus, the photon it emits during this process will always have the same wavelength. For the hydrogen atom, the energy difference between the n = 2 and n = 1 energy levels is 10.19eV, corresponding to a photon wavelength of \(\lambda\) = 1215.67 Angstroms. This is the most commonly-observed atomic transition in all of astronomy, as hydrogen is by far the most abundant element in the Universe. It is referred to as the Lyman α transition (or Lyα for short).
It turns out that the strongest emission feature you observed in the quasar spectrum above arises from Lyα emission from material orbiting around the central black hole.
a) Recall the Doppler equation:
\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z\approx\frac{v}{c}\]
Using the data provided, calculate the redshift of this quasar.
Looking at the spectrum, the peak appears to be at about 1405 Angstroms. Since we know the actual wavelength, we can just plug the numbers in to calculate the redshift.
\(z=\frac{1405Å-1216Å}{1216Å}=0.155\)
b) Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It will help to know that the typical accretion disk around a \(10^8M_\odot\) black hole extends to a radius of \(r=10^{15}m\).
The Virial Theorem can be used to show that \(M\approx\frac{\sigma^2R}{G}\). We then need to find \(\sigma\) (the stellar velocity dispersion) so that we can calculate the mass. We can do this by looking at the wavelength at half the peak flux (which appears to be about 1400Å and 1415Å). Taking the average difference between these value and the peak wavelength (1405Å) gives us 7.5Å. We can divide this wavelength by the peak wavelength to calculate the dispersion.
\(\frac{7.5Å}{1405Å}=0.0053\approx\frac{\sigma}{c}\)
\(\sigma=0.0053*c=1.59*10^6m/s\)
Finally, we can plug this into the equation, using the typical radius.
\(M=\frac{(1.59*10^6m/s)^2(10^{15}m)}{6.674*10^{-14}m^3g^{-1}s^{-2}}=3.79*10^{40}g\rightarrow1.90*10^7M_\odot\)
Careful! The denominator of your velocity dispersion equation ought to be the *emitted* wavelength rather than the shifted one! Making this adjustment, you might arrive at a slightly larger black hole mass!
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