# Using emission lines to determine redshift of a quasar

I am attempting this past paper question and am not sure how to tackle this, this is not homework! :)

The spectrum of a distant quasar shows two broad emission lines with observed wavelengths of 317.7nm and 404.7nm. The strongest lines are most likely to be either Lyman-α with a rest wavelength 121.6nm, CIV with a rest wavelength of 154.9nm or MgII with a rest wavelength of 280.0nm.

(a) What is the redshift of the quasar? Please show your working.

(b) What two lines have been detected?

(c) The two lines have widths of 10nm and 20nm respectively. If I have measured the mass of the black hole in the quasar to be 10^9Msun, how far from the centre of the quasar are the regions of ionised gas responsible for the emission lines? Please specify any assumptions that you make to obtain your result.

I am assuming I must find z to be the ratio of the change in wavelength to the rest wavelength. However I'm unsure of how to identify these in the question.

Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman $$\alpha$$ and C IV, but it's not true that the two obtained redshifts are suspiciously alike. In fact, when you do the calculation, you get (from $$z \equiv \lambda_\mathrm{obs}/\lambda_\mathrm{rest} - 1$$), $$\begin{array}{rcl} z_{\mathrm{Ly}\alpha} & = & \frac{317.7\,\mathrm{nm}}{125.67\,\mathrm{nm}} & \simeq & 1.6134 \\ z_\mathrm{C\,IV} & = & \frac{404.7\,\mathrm{nm}}{154.9\,\mathrm{nm}} & \simeq & 1.6127. \end{array}$$ That is, Lyman $$\alpha$$ is slightly more redshifted than C IV. This is often seen, and is due to Lyman $$\alpha$$ scattering resonantly on the neutral hydrogen enshrouding the quasar / host galaxy. This effect is more pronounced the higher the redshift, and by $$z\simeq6$$, the blue part of the spectrum is completely gone (this is the so-called Gunn-Peterson trough).
Relating the widths $$\Delta\lambda$$ of the lines to the motion around the black hole, the velocity of the gas responsible for emitting the lines is given by $$v/c = \Delta\lambda / \lambda$$, so the velocities are $$\begin{array}{rcl} v_{\mathrm{Ly}\alpha} & = & c \frac{10\,\mathrm{nm}}{317.7\,\mathrm{nm}} & \simeq & 9\,400\,\mathrm{km}\,\mathrm{s}^{-1} \\ v_\mathrm{C\,IV} & = & c \frac{20\,\mathrm{nm}}{404.7\,\mathrm{nm}} & \simeq & 14\,800\,\mathrm{km}\,\mathrm{s}^{-1}. \end{array}$$ From simple dynamics, this can be converted to a distance $$r$$ from the black hole, knowing its mass $$M_\bullet$$: $$\begin{array}{rcl} r & = & \frac{G M_\bullet}{v^2} & \Rightarrow \\ r_{\mathrm{Ly}\alpha} & \simeq & 0.05\,\mathrm{pc} & \simeq & 58\,\mathrm{light\text{-}days} \\ r_{\mathrm{C\,IV}} & \simeq & 0.02\,\mathrm{pc} & \simeq & 23\,\mathrm{light\text{-}days}, \end{array}$$ which is consistent with the fact that these lines are formed in the so-called broad line region which has dimension from tens of light-days to several parsec (in contrast, the narrow line region has dimensions of several 100s to ~1000 light-years).
So the two observed lines must be two of the three suggested lines, red-shifted by the same amount. That means the ratio of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio $$404.7/317.6$$. It's easy to check that only the first two are close to that ratio, so they must be those two. So the redshift should be equal to $$317.7/121.6 = 2.61$$ and to $$404.7/154.9$$ which is also $$2.61$$ (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.
• You forgot to subtract $1$ from the ratio, so the redshift is only 1.61. More precisely, $z_{\mathrm{Ly}\alpha} = 1.6134$ and $z_{\mathrm{C\,IV}} = 1.6127$, so Lyα is a bit more redshifted than C IV which is not unusual, and may be expected due to to resonant scattering. – pela Jun 10 at 17:52