A Strömgren sphere and neutral time

[TFM]

Homework Statement

A Strömgren sphere is the ionized region surrounding a strong ionizing source. Its size is determined by there being a balance between the rate of ionization and the rate of recombination. Consider such a sphere of radius $$R_S$$ = 10 pc and internal density $$n_e = 10^6 m^{-3}$$, and with a central ionizing source of $$10^{49} photons s^{−1}$$.

a)

How long will it take to become neutral once the ionising star has switched off?

b)

Suppose that the sphere expands at a rate equal to the sound speed in a 10^4 K gas. How long will it take to expand to a radius of 100 pc? ($$v_s^2\approx 3kT/m_H$$)

Homework Equations

The Attempt at a Solution

Okay I have done part b), getting an answer of about 5.6 million years, but I have a small query regarding part a).

I know that the charge loss will be $$10^49 atoms s^{-1}$$
because this is the photon rate when it is in equilibrium, which means this rate is equal to the rate at which atoms become neutral. I also know that I need to work out the volume of the sphere, and thus the number of particles in it.

However, i am slightly uncertain how to work out how many particles are actually ionised before the star is turned off.

?

Thanks in advanced,

TFM

 

[nate808]

Dear TFM,

Thank you for your post. Let me help you with part a) of your question.

To calculate the time it takes for the Strömgren sphere to become neutral once the ionizing star has switched off, we need to consider the balance between the rate of ionization and the rate of recombination. In equilibrium, these two rates are equal, which means that the number of ionizations and recombinations happening per unit time are the same.

As you correctly stated, the rate of ionization is given by 10^{49} photons s^{-1}. To calculate the rate of recombination, we need to consider the number of particles in the sphere and their recombination rate.

The number of particles in the sphere can be calculated using the volume of the sphere, which is given by V = \frac{4}{3}\pi R_S^3. This volume is filled with particles at a density of n_e = 10^6 m^{-3}, so the number of particles in the sphere is given by N = n_eV = \frac{4}{3}\pi R_S^3 n_e.

Next, we need to consider the recombination rate of these particles. Recombination is a process in which an ion and an electron combine to form a neutral atom. The recombination rate is given by the product of the number of ions and the electron density, which is equal to N n_e = \frac{4}{3}\pi R_S^3 n_e^2.

Now, if we equate the rate of ionization with the rate of recombination, we get the following equation:

10^{49} = \frac{4}{3}\pi R_S^3 n_e^2

Solving for n_e, we get n_e = \sqrt{\frac{3\times10^{49}}{4\pi R_S^3}}.

Substituting the values for R_S and n_e, we get n_e \approx 1.6 \times 10^5 m^{-3}.

This means that at equilibrium, there are approximately 1.6 \times 10^5 ions in the sphere. However, once the star is switched off, the rate of ionization becomes zero, while the rate of recombination remains the same. This means that the number of ions will decrease over time until the sphere becomes neutral.

To calculate the time it takes for the sphere to

 

[Blueshift5]

Hello TFM,

To calculate the number of particles that are ionized before the star is turned off, we can use the fact that the rate of ionization is equal to the rate of recombination when the Strömgren sphere is in equilibrium. This means that the number of ionizations per second is equal to the number of recombinations per second. We can use this to calculate the number of ionized particles in the sphere at any given time.

For part a), we can use the given information that the central ionizing source emits 10^{49} photons per second and the sphere has a radius of 10 pc. This means that the photon flux at the surface of the sphere is 10^{41} photons per second per square meter (using the inverse square law). We can then use the formula for the rate of ionization, which is given by:

N_{ion} = \alpha n_e^2 V

where N_{ion} is the number of ionizations per second, \alpha is the recombination coefficient, n_e is the electron density, and V is the volume of the sphere. We can rearrange this formula to solve for V, which gives:

V = \frac{N_{ion}}{\alpha n_e^2}

We know that the rate of ionization is equal to the photon flux at the surface of the sphere, so we can substitute that in for N_{ion}. We also know the electron density, n_e, from the given information. The recombination coefficient, \alpha, can vary depending on the temperature and composition of the gas, but for a typical value we can use \alpha = 2.6 \times 10^{-19} m^3 s^{-1}. Plugging in these values, we get:

V = \frac{10^{41} \text{ photons m}^{-2} \text{s}^{-1}}{2.6 \times 10^{-19} \text{ m}^3 \text{s}^{-1} \times (10^6 \text{ m}^{-3})^2} \approx 1.5 \times 10^{14} \text{ m}^3

This is the volume of the sphere when it is in equilibrium, which is also the number of particles that are ionized before the star is turned off. We can then use this to calculate the time it takes for the sphere to