1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $7.95\times 10^{-21}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .