Use the Saha equation to determine the fraction of hydrogen atoms
that are ionised, NII/Ntot, at the centre of the Sun. Here the temperature is 15.7 × 10^6 K
and the number density of electrons is about ne = 6.1 × 10^31 m−3. (Use ZI = 2, and ZII = 1,
since ionised H has only one state).

The Saha equation is used to determine the ratio of ionized to neutral atoms in a gas, based on temperature, electron density, and partition functions. The equation can be written as:

\[
\frac{N_{\text{II}}}{N_{\text{I}}} = \frac{g_{\text{II}}}{g_{\text{I}}} \left(\frac{2 \pi m_e k_B T}{h^2}\right)^{3/2} \frac{1}{n_e} e^{-\frac{\chi}{k_B T}}
\]

where:
- \( \frac{N_{\text{II}}}{N_{\text{I}}} \) is the ratio of ionized to neutral hydrogen atoms,
- \( g_{\text{II}} \) and \( g_{\text{I}} \) are the statistical weights (partition functions) for ionized and neutral hydrogen,
- \( m_e \) is the mass of the electron (\(9.109 \times 10^{-31} \, \text{kg}\)),
- \( k_B \) is Boltzmann’s constant (\(1.381 \times 10^{-23} \, \text{J/K}\)),
- \( T \) is the temperature in Kelvin,
- \( h \) is Planck’s constant (\(6.626 \times 10^{-34} \, \text{J·s}\)),
- \( n_e \) is the electron number density, and
- \( \chi \) is the ionization energy of hydrogen (13.6 eV or \( 2.179 \times 10^{-18} \, \text{J} \)).

For hydrogen, the partition functions are \( Z_I = 2 \) and \( Z_{\text{II}} = 1 \). Additionally, the ratio \( \frac{N_{\text{II}}}{N_{\text{tot}}} \) can be written as:

\[
\frac{N_{\text{II}}}{N_{\text{tot}}} = \frac{\frac{N_{\text{II}}}{N_{\text{I}}}}{1 + \frac{N_{\text{II}}}{N_{\text{I}}}}
\]

Now, let's compute the fraction of ionized hydrogen at the center of the Sun, using the given values.

Given:
- \( T = 15.7 \times 10^6 \, \text{K} \),
- \( n_e = 6.1 \times 10^{31} \, \text{m}^{-3} \),
- \( Z_I = 2 \),
- \( Z_{\text{II}} = 1 \),
- Ionization energy \( \chi = 13.6 \, \text{eV} \).

We will now calculate the ratio \( \frac{N_{\text{II}}}{N_{\text{I}}} \) and then use it to find the fraction \( \frac{N_{\text{II}}}{N_{\text{tot}}} \).

I'll compute this step-by-step.The fraction of hydrogen atoms that are ionized at the center of the Sun, \( \frac{N_{\text{II}}}{N_{\text{tot}}} \), is approximately 0.549. This means that around 54.9% of the hydrogen atoms are ionized at the Sun's core.

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How does the electron density affect the ionization fraction in the Saha equation?
2. What happens to the ionization fraction if the temperature of the Sun’s core changes?
3. How does the Saha equation apply to other elements besides hydrogen?
4. Can the Saha equation be used for non-thermal equilibrium situations?
5. What is the significance of the partition function in the Saha equation?

**Tip:** The Saha equation is particularly useful in astrophysics for estimating ionization levels in stars based on temperature and density.

Use the Saha equation to determine the fraction of hydrogen atoms that are ionized, NII/Ntot, at the centre of the Sun. Here the temperature is 15.7 × 10^6 K and the number density of electrons is about ne = 6.1 × 10^31 m−3. (Use ZI = 2, and ZII = 1, since ionised H has only one state).

Learn how to use the Saha equation to calculate the ionization fraction of hydrogen atoms at the Sun's core with a temperature of 15.7 × 10^6 K and an electron density of 6.1 × 10^31 m−3. This astrophysics problem covers key concepts in thermodynamics and statistical mechanics.

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