Blog Post 24, WS 7.2, Problem 4: Quasar Mass

4. One feature you surely noticed was the strong, broad emission lines. Here is a closer look at the strongest emission line in the spectrum:

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, labeled n = 1 in the model of the hydrogen atom shown below (and called the ground state): It turns out that that strongest emission feature you observed in the quasar spectrum above arises from Lyα emission from material orbiting around the central black hole.

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, and so only the n = 2 and n = 3 excited states are shown above for simplicity.

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 10.19 eV, corresponding to a photon wavelength of λ = 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).

(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.

(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 108 M⊙ black hole extends to a radius of r = 1015 m.

OK, let's begin.

(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. 

So, the rest wavelength that we are observing is 1215.67\(\unicode[serif]{xC5}\).  However, the intensity in the data peaks at the wavelength 1407 \(\unicode[serif]{xC5}\).  Let's use the equation:

\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z \approx \frac{v}{c}\]

\[\frac{1407-1215.67}{1215.67}=z \approx \frac{v}{c}\]

\[\boxed{z=0.157\approx \frac{v}{c}}\]

(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 108 M⊙ black hole extends to a radius of r = \(10^{15}\) m.

This relies on a few different equations.  We need our Doppler that was used in part (a), but we also need Virial Theorem solved for mass.

\[K=-\frac{1}{2}U\]

\[Mv^2=\frac{3GM^2}{5R}\]

\[M=\frac{5v^2 R}{3G}\]

We are given R, but we now need V.  For this, we will use the Doppler data.

We know the approximate speed that the black hole is traveling away from us: 0.157c.  If we repeat this calculate for the matter at the edge of the spectrum spike, then we can get the speed of the outer edge matter.  By subtracting the two, we can get the relative speed of the outside matter.

\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z \approx \frac{v}{c}\]

\[\frac{1395-1215.67}{1215.67}=z \approx \frac{v}{c}\]

\[\frac{v}{c}=0.148\]

Now let's compare the two:

\[\Delta v=v_{BH}-v_1=0.157c-0.148c=0.009c=2.7\times 10^8cm/s\]

Let's plug this into our final equation:

\[M=\frac{5v^2 R}{3G}\]

\[M=\frac{5(2.7\times 10^8cm/s)^2 (10^{17 cm})}{3(6.7\times 10^{-8}cm^3 g^{-1}s^{-2}}\]

\[\boxed{M_{BH}=1.8\times 10^{41}g=9.1\times 10^7 M_{\odot}}\]

I worked with B. Brzycki, G. Grell, and N. James.