SAHA’S IONIZATION FORMULA NAME:- ABHISHEK & UDDESHYA

Astronomers wanted to know “WHAT IS SUN MADE OF ?” IN 19TH Century Physicist said “LOOK AROUND US, EARTH IS MADE OF HEAVY ELEMENTS. SO SUN IS NOT VERY DIFFERENT IT IS ALSO MADE OF HEAVY ELEMENTS LIKE COPPER IRON OR WHATEVER IT IS.” But we all know that sun is made of mostly Hydrogen. How did we come to this conclusion.

If you heat up something, it glows. It change the colors If you heat up something, it glows .It change the colors. So SAHA thought that if it can happen in laboratory it can also happen stars and in whole universe. The idea is if you excite an atom its electron goes to higher state and during returning to normal state it emits the radiation. From there to think stars temperature can be calculated by its line emission. It was an extra ordinary Idea.

MEGHNAAD SAHA, HE was the first person who Applied quantum theory on stellar objects He wrote a paper titled “IONIZATION IN SOLAR CHROMOSPHERE.” And it was published in PHIL MAG in 1920 while he In Europe.

There was a young lady in HARVARD “Henrietta Leavitt” and she said let me use SAHA’s new equation and try to estimate how much hydrogen is in the sun. Nobody has done that before and then she said sun is made of mostly hydrogen. Her thesis examiner was high priest of Astrophysics, Henry Norris Russel professor of astrophysics in Princeton University. He sat with Saha’s papers and check and instead of hushing it up, he openly came and said that I was wrong we all were wrong she is correct , Starts and sun are made of Hydrogen. Henrietta Leavitt Henry Norris Russel

SAHA IONIZATION FORMULA Saha's ionization formula:- following Saha , we recall that in the interior of stars, temperatures are ,extremely high and the elements present there are mostly in the atomic state. argued that under the prevailing conditions, atoms move very rapidly fid undergo frequent collisions. In the process, they are stripped of valence electrons This is referred to as thermal ionization and is accompanied by electron recapture to form neutral atoms. Saha's ionization formula relates the temperature, pressure and ionization potential of atoms to their degree of ionization.

(a) ------(1) (a) the contribution of nuclear spin is negligible, (b) only single ionization occurs, and (c) dynamical equilibrium between ionization and recombination (due to electron capture) is reached at a given temperature and pressure, so that (a) ------(1)

where A+ is a singly ionized positive ion. This may be treated as a reaction between different components of an ideal gas so that at equilibrium (b) ...........(2) where k is the component index for A, A+ or is the stoichiometric coefficient* of the kth component, and is the chemical potential of the kth atom. (c)---------(3)

(e) -----(5) N partical partition function can be written as On combining Eqs (3) and (4), we find that the chemical potential is given by (e) -----(5)

where Z (T) refers to the contributions of the internal modes (rotational, vibrational and electronic) to the partition function. Note that it is a function Of T but not of pressure. This configuration is equivalent to the presence of several gases contained in a vessel. Let us assume that pa, pi and pe are the partial pressures exerted by the atom, the ion and the electron respectively. Let their masses be ma, and me, respectively. Hence, the chemical potential of the kth component in the System is given by NA - Avogadro number. It is important to point out here that for most reactions of the type equation (1) we measure energies from the ground state. The partition functions in this case are given by

(g) where €* is the ionisation energy, At equilibrium Eq. (2) implies that (h) Assumption (b) implies that Ve = Vi = Va = 1. substituting for the chemical potentials with appropriate partition functions, we get (i)

(j)-------(7) ■ On rearranging terms, we can write The ratio Mi/Ma = 1, as the mass of a hydrogen atom is 10000 times the mass of an electron. Further, for free atoms and ions, the possible Internal states are electronic states. For all practical purposes, these may be taken to be the ground state degeneracies ga, gi and ge. For an electron, ge = 2 and me =Me/Na Then, we can rewrite Eq. (7) as

and pa = p -2pe (K)-------(8) (m)------(9) Hence, Where x = (pe/p) denotes the mole fraction of electrons, i.e. the fraction of electrons in the system. On combining Eqs (8) and (9), we get pa = p -2pe Hence, (m)------(9)

(n)-------(10) This equation expresses the degree of ionization as a function of temperature, pressure and the ionization potential of the atoms present in the interior of a star. This is commonly referred to as Saha's ionization formula. It implies that the degree of ionization will be more if temperature is high, or pressure and ionization potential are low. This formula finds many applications in astrophysics. It has been successfully used to explain the Fraunhofer spectrum of the sun, and the spectrum of stars, and to estimate the temperature of the atmosphere. Now a days it is finding wide applications in plasma studies, as also in the ionization processes in semi- conductors.

(o)-------(11) pa = pe = pi H ⇌ p+ + e- We can illustrate the use of Eq. (10) by considering the ionisation of a hydrogen atom : H ⇌ p+ + e- In this reaction the process proceeds from the sta te of predominantly neutral hydrogen to predominantly proton. Hence, we may assume pa = pe = pi so that x = (1/3), the ionisation energy of hydrogen is 13.6eV. The proton has two spin states, i.e. & gi = 2 and for the hydrogen atom ga = 4,because of 2 electron and 2 proton spin states. Let the density of all the particles be n. Then, p = nkBT. Hence, from Eq. (10), we get (o)-------(11)

(P) Variation of n as a function of T for the ionisation equilibrium of hydrogen. Since one has to go for very high , the logarithmic plot, rather than a linear plot is more convenient

“The relation between n and T based on Eq “The relation between n and T based on Eq. (11) is shown in Fig(p) neutral hydrogen dominates the radiation at high atomic densities and low temperatures while the protons and electrons dominate at high temperatures and low densities. Let us take n = 1 cm-3 and n = 1025 cm-3 . Then from Fig(p) , we observe that the corresponding temperatures are given by log10 T= 3.8 and log10 T= 6. This means that the transition occurs at 1038 K (7500 K) and 106 K or n = 1 cm-3 and n = 1025 cm-3, respectively. “

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