Question

Galaxies tend to be strong emitters of Lyman $\mathit{\alpha }$photons(from the n=2 to n=1 transition in atomic hydrogen). But theintergalactic medium—the very thin gas between the galaxies—tends to absorb Lyman$\mathit{\alpha }$photons. What can you infer from theseobservations about the temperature in these two environments?

Short answer

The temperature is$T=23.9\times {10}^{3}K$

Step-by-step solution

Important Concepts

For black bodies we have wein’s displacement law, stating that peaks of temperature at wavelengths that are inversely proportional to temperature

${\mathbf{\lambda }}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{90}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mk}}{\mathbf{T}}$

Where T is the temperature and${\mathit{\lambda }}_{\mathbf{l}\mathbf{n}}$ is the maximum wavelength at that temperature.

For electronic transition we have

role="math" localid="1663931759430" $\frac{\mathbf{1}}{{\mathbf{\lambda }}_{\mathbf{m}}}\mathbf{=}\mathit{R}\mathbf{(}\frac{\mathbf{1}}{{\mathbf{n}}_{\mathbf{1}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}}\mathbf{)}$

Where R is the Rydberg constant and n₁and n₂ are the transition shells.

Application

Find the wavelength being emitted by the atoms in the galaxy

We have

$\frac{1}{{\mathrm{\lambda }}_{\mathrm{m}}}=R(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2})$

Input the values we get

$$\begin{gathered} \frac{1}{{\lambda }_m}=\left(1.097\times {10}^7{m}^{-1}\right)(\frac{1}{4}-\frac{1}{4}) \\ \frac{1}{{\lambda }_m}=82227500m^{-1} \\ {\lambda }_m=1.22\times {10}^{-7}m \end{gathered}$$

Considering galaxies to be black bodies , input this into the wein’s displacement law

$$\begin{gathered} T=\frac{2.90\times {10}^{-3}mk}{1.22\times {10}^{-7}m} \\ T=23.9\times {10}^{3}K \end{gathered}$$