Math Problem Statement

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).

Solution

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:

NIINI=gIIgI(2πmekBTh2)3/21neeχkBT\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:

  • NIINI\frac{N_{\text{II}}}{N_{\text{I}}} is the ratio of ionized to neutral hydrogen atoms,
  • gIIg_{\text{II}} and gIg_{\text{I}} are the statistical weights (partition functions) for ionized and neutral hydrogen,
  • mem_e is the mass of the electron (9.109×1031kg9.109 \times 10^{-31} \, \text{kg}),
  • kBk_B is Boltzmann’s constant (1.381×1023J/K1.381 \times 10^{-23} \, \text{J/K}),
  • TT is the temperature in Kelvin,
  • hh is Planck’s constant (6.626×1034J\cdotps6.626 \times 10^{-34} \, \text{J·s}),
  • nen_e is the electron number density, and
  • χ\chi is the ionization energy of hydrogen (13.6 eV or 2.179×1018J2.179 \times 10^{-18} \, \text{J}).

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

NIINtot=NIINI1+NIINI\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=15.7×106KT = 15.7 \times 10^6 \, \text{K},
  • ne=6.1×1031m3n_e = 6.1 \times 10^{31} \, \text{m}^{-3},
  • ZI=2Z_I = 2,
  • ZII=1Z_{\text{II}} = 1,
  • Ionization energy χ=13.6eV\chi = 13.6 \, \text{eV}.

We will now calculate the ratio NIINI\frac{N_{\text{II}}}{N_{\text{I}}} and then use it to find the fraction NIINtot\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, NIINtot\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Astrophysics
Statistical Mechanics
Thermodynamics

Formulas

Saha equation: (N_II / N_I) = (g_II / g_I) * (2πm_e k_B T / h^2)^(3/2) * (1/n_e) * e^(-χ / k_B T)
Fraction of ionized atoms: N_II / N_tot = (N_II / N_I) / (1 + N_II / N_I)

Theorems

Saha Ionization Equation

Suitable Grade Level

Undergraduate level (Physics, Astronomy)