<p>Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman <span class="math-container">$\alpha$</span> 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 <span class="math-container">$z \equiv \lambda_\mathrm{obs}/\lambda_\mathrm{rest} - 1$</span>),
<span class="math-container">$$
\begin{array}{rcl}
z_{\mathrm{Ly}\alpha} &amp; = &amp; \frac{317.7\,\mathrm{nm}}{125.67\,\mathrm{nm}} &amp; \simeq &amp; 1.6134 \\
z_\mathrm{C\,IV}      &amp; = &amp; \frac{404.7\,\mathrm{nm}}{154.9\,\mathrm{nm}} &amp; \simeq &amp; 1.6127.
\end{array}
$$</span>
That is, Lyman <span class="math-container">$\alpha$</span> is slightly more redshifted than C IV. This is often seen, and is due to Lyman <span class="math-container">$\alpha$</span> scattering resonantly on the neutral hydrogen enshrouding the quasar / host galaxy. This effect is more pronounced the higher the redshift, and by <span class="math-container">$z\simeq6$</span>, the blue part of the spectrum is completely gone (this is the so-called <a href="https://en.wikipedia.org/wiki/Gunn%E2%80%93Peterson_trough" rel="nofollow noreferrer">Gunn-Peterson trough</a>).</p>

<p>Relating the widths <span class="math-container">$\Delta\lambda$</span> of the lines to the motion around the black hole, the velocity of the gas responsible for emitting the lines is given by <span class="math-container">$v/c = \Delta\lambda / \lambda$</span>, so the velocities are
<span class="math-container">$$
\begin{array}{rcl}
v_{\mathrm{Ly}\alpha} &amp; = &amp; c \frac{10\,\mathrm{nm}}{317.7\,\mathrm{nm}} &amp; \simeq &amp; 9\,400\,\mathrm{km}\,\mathrm{s}^{-1} \\
v_\mathrm{C\,IV}      &amp; = &amp; c \frac{20\,\mathrm{nm}}{404.7\,\mathrm{nm}} &amp; \simeq &amp; 14\,800\,\mathrm{km}\,\mathrm{s}^{-1}.
\end{array}
$$</span>
From simple dynamics, this can be converted to a distance <span class="math-container">$r$</span> from the black hole, knowing its mass <span class="math-container">$M_\bullet$</span>:
<span class="math-container">$$
\begin{array}{rcl}
r &amp; = &amp; \frac{G M_\bullet}{v^2} &amp; \Rightarrow \\
r_{\mathrm{Ly}\alpha} &amp; \simeq &amp; 0.05\,\mathrm{pc} &amp; \simeq &amp; 58\,\mathrm{light\text{-}days} \\
r_{\mathrm{C\,IV}} &amp; \simeq &amp; 0.02\,\mathrm{pc} &amp; \simeq &amp; 23\,\mathrm{light\text{-}days},
\end{array}
$$</span>
which is consistent with the fact that these lines are formed in the so-called <em>broad line region</em> 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).</p>

CC BY-SA 4.0
2019-06-10T22:55:54.480Z
2019-06-10T22:55:54.480Z
pela
<p>So the two observed lines must be two of the three suggested lines, red-shifted by <strong>the same amount</strong>. That means the <em>ratio</em> of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio <span class="math-container">$404.7/317.6$</span>.  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 <span class="math-container">$317.7/121.6 = 2.61$</span>  and to <span class="math-container">$404.7/154.9$</span> which is also <span class="math-container">$2.61$</span> (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.</p>

CC BY-SA 4.0
2019-06-10T17:52:17.777Z
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
CC BY-SA 4.0
2019-06-10T11:08:31.623Z
2019-06-10T11:08:31.623Z
Steve Linton
<p>I am attempting this past paper question and am not sure how to tackle this, this is not homework! :)</p>

<blockquote>
  <p>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.</p>
  
  <p>(a) What is the redshift of the quasar? Please show your working.</p>
  
  <p>(b) What two lines have been detected?</p>
  
  <p>(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.</p>
</blockquote>

<p>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.</p>

<redshift><wavelength><spectra>
Using emission lines to determine redshift of a quasar
Minish
# Using emission lines to determine redshift of a quasar

(Asked by: username_0 on 2019-06-10)

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.

**Tags:** redshift, wavelength, spectra

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## 2 answers

### Answer by username_1 on 2019-06-10. Score: 2 | Accepted answer

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](https://en.wikipedia.org/wiki/Gunn%E2%80%93Peterson_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).


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### Answer by username_2 on 2019-06-10. Score: 4

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.

> **Comments:**
> **username_1**: 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.  

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