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

• LCSphysicist
In summary, the number of particles per unit volume after ionisation of hydrogen molecules is equal to 2 times the density divided by the mass of a single hydrogen atom, rather than the density divided by half the mass of a single hydrogen atom. This is because each hydrogen molecule, when fully ionised, produces 4 particles, 2 protons and 2 electrons, so the total number of particles is twice the number of molecules.
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
.

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?

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##

Herculi said:
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.)

LCSphysicist

## 1. What is black body radiation?

Black body radiation is the electromagnetic radiation emitted by an object that absorbs all the radiation that falls on it. It is also known as thermal radiation or Planck radiation.

## 2. What is the relationship between temperature and black body radiation?

The intensity and wavelength distribution of black body radiation are directly related to the temperature of the object. As the temperature increases, the intensity of radiation also increases and the peak wavelength shifts to shorter wavelengths.

## 3. How does gas affect black body radiation?

Gases can absorb and emit specific wavelengths of radiation, which can affect the overall intensity and distribution of black body radiation. This is known as the absorption and emission spectrum of the gas.

## 4. What is the significance of Wien's displacement law in black body radiation?

Wien's displacement law states that the peak wavelength of black body radiation is inversely proportional to the temperature of the object. This law is important in understanding the relationship between temperature and the intensity of radiation.

## 5. How is black body radiation related to the study of thermodynamics?

Black body radiation is a fundamental concept in thermodynamics, as it is an important part of understanding the transfer of heat and energy between objects. It also plays a crucial role in the study of the laws of thermodynamics and the behavior of gases.

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