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):
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 $n = 3$ 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 10.19 eV, 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 $\alpha$ transition (or Ly$\alpha$ for short).
It turns out that that strongest emission feature you observed in the quasar spectrum aboves arises from Ly$\alpha$ emission from material orbiting around the central black hole.
(a) Recall the Doppler equation:
\begin{align*}
\frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = z \approx \frac{v}{c}
\end{align*}
Using the data provided, calculate the redshift of this quasar.
Looking at the zoomed-in graph of the spectrum above, the peak of the emission line is at 1410 Angstroms, which shows that the $\lambda_{observed} = 1410$ Angstrom. We know that the actual emitted emission for the Ly$\alpha$ transition is $\lambda_{emitted} = 1215.67$. Using the equation above, we can solve for the redshift as follows:
\begin{align*}
z &= \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}\\
z &= \frac{1410 - 1215.67}{1215.67}\\
z &\approx 0.16
\end{align*}
Therefore, the redshift of this quasar is 0.16.
(b) Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It would help to know that the typical accretion disk around a $10^8 M_{\odot}$ black hole extends to a radius of $r = 10^{15}$ m.
The velocity at which this quasar is moving away from us can be found by looking at the width of the broadened peak. The broadened peak is about 15 Angstroms, so the redshift between these two peaks would be $z = \frac{15}{1215} \approx 0.012$.
However, we only use half that value for the actual velocity, since half of the quasar is coming towards us, and the other half is going away from us. Therefore, $z = 0.006$. Using the equation above we can calculate how fast the quasar is going away from us:
\begin{align*}
z &= \frac{v}{c}\\
v&= zc\\
v &= 0.006c
\end{align*}
Now that we have the speed at which the quasar is moving away from us, we can use the Virial Theorem to solve for the mass of the black hole, $M$ as follows:
\begin{align*}
K &= -\frac{1}{2}U\\
\frac{1}{2}mv^2 &= \frac{1}{2}\frac{GMm}{r}\\
v^2 &= \frac{GM}{r}\\
M &= \frac{v^2r}{G}\\
M &= \frac{(0.006 \times 3 \times 10^{10})^2(10^{17})}{6.7 \times 10^{-8}}\\
M &= 4.8 \times 10^{40} \text{ grams}\\
M &= 2.4 \times 10^7 M_{\odot}
\end{align*}
The mass of the black hole is $2.4 \times 10^7 M_{\odot}$.
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