What Determines the Number of Particles per Unit Volume in Ionized Hydrogen?

[LCSphysicist]

Homework Statement

Find the temperature of totally ionized hydrogen plasma
of density p = 0.10 g/cm^ at which the thermal radiation pressure
is equal to the gas kinetic pressure of the particles of plasma. Take
into account that the thermal radiation pressure p = u/3, where u
is the space density of radiation energy, and at high temperatures all
substances obey the equation of state of an ideal gas.

Relevant Equations

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The image above is the solution posted by the book. I can follow the reasoning that has been used, but i have a trouble particularly at the first equation itself.

Why should $$n = 2 \rho / m_{H} $$ instead of $$n = \rho / (2m_{H})$$, since the mass os a molecule of hydrogen is two times the hydrogen/proton mass?

 

[Bandersnatch]

If you had only protons there, the mass would be the same, and the density would be the same. From $n=\rho / m_H$ you'd get a number of protons. Now, since you know there is an extra particle for each proton that doesn't really do anything mass-wise, you get twice the number of particles that the same density/mass consideration would suggest. Unsurprisingly, $n_{proton+electron}=2n_{proton}$.

The equation you think should be there instead suggests that you should only count half as many particles as what you get from density/mass considerations.

You could imagine some fantasy physics where each proton is accompanied by not only an electron, but also an additional imaginary particle that doesn't do anything mass-wise. The number of particles $n$ in that case should then be 3 times what you get from density/mass. Which would be $n=3\rho / m_H$

 

[Steve4Physics]

Why should $$n = 2 \rho / m_{H} $$ instead of $$n = \rho / (2m_{H})$$, since the mass os a molecule of hydrogen is two times the hydrogen/proton mass?

The number of hydrogen molecules (each of mass $2m_H$) per unit volume is $\frac {\rho}{2m_H}$.

One hydrogen molecule, when fully ionised, produces 4 particles (2 protons and 2 electrons).

So, after ionisation, the number of particles per unit volume is $4 \times \frac {\rho}{2m_H} = \frac {2\rho}{m_H}$.

(Note, it’s probably better to think in terms of atoms rather than molecules for these sorts of problems.)