<p>The mass of a hydrogen atom is <span class="math-container">$1.67\times 10^{-27}$</span> kg <span class="math-container">$\simeq 8.35\times 10^{-58} M_\odot$</span>. A cm is <span class="math-container">$3.24 \times 10^{-19}$</span> pc. The conversion factor is that the column density needs to be multiplied by  <span class="math-container">$8.35\times 10^{-58}$</span> and divided by <span class="math-container">$(3.24 \times 10^{-19})^2$</span> to put it in units of <span class="math-container">$M_\odot/$</span>pc<span class="math-container">$^2$</span>.</p>
<p>i.e. Multiply by <span class="math-container">$7.95\times 10^{-21}$</span>.</p>

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2023-07-04T13:15:05.053Z
2023-07-04T13:15:05.053Z
ProfRob
<p>I know the column density distribution of neutral hydrogen gas (units: cm^-2). I want to calculate the corresponding surface density distribution of the neutral hydrogen gas (units: M_solar/pc^2). How can I achieve the same?</p>

<galaxy><radio-astronomy>
What is the relation between column density and surface density of neutral hydrogen gas HI?
SlightDecoy
# What is the relation between column density and surface density of neutral hydrogen gas HI?

(Asked by: username_0 on 2023-07-04)

I know the column density distribution of neutral hydrogen gas (units: cm^-2). I want to calculate the corresponding surface density distribution of the neutral hydrogen gas (units: M\_solar/pc^2). How can I achieve the same?

**Tags:** galaxy, radio-astronomy

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## 1 answers

### Answer by username_1 on 2023-07-04. Score: 1

The mass of a hydrogen atom is $1.67\times 10^{-27}$ kg $\simeq 8.35\times 10^{-58} M\_\odot$. A cm is $3.24 \times 10^{-19}$ pc. The conversion factor is that the column density needs to be multiplied by $8.35\times 10^{-58}$ and divided by $(3.24 \times 10^{-19})^2$ to put it in units of $M\_\odot/$pc$^2$.

i.e. Multiply by $7.95\times 10^{-21}$.


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