# Hydrogen in Sun

Inside the Sun the average temperature is $10^7 K$. For Hydrogen atoms the ground state $E_1=-13.6 eV$ (1S)  has degeneracy one, and the first excited state has degeneracy 4 $E_2=-13.6/4=-3.4 eV$ (S and P(x,y,z)). Please find the ratio of Hydrogen atoms that are in the first excited state to the Hydrogen atoms that are in the ground state.

There are possible 5 states, one ground state of energy $E_1$ and 4 excited states of energy $E_2$. If we write the sum of all states (the partition function from statistical physics):

$Z=\exp(-E_1/kT)+4\exp(-E_2/kT)$

then the probability of having energy $E_1$ is (according to the classical Boltzmann distribution):

$P_1=\frac{\exp(-E_1/kT)}{Z}$

The probability of having energy $E_2$ is (4 is the number of states having energy $E_2$):

$P_2=\frac{4\exp(-E_2/kT)}{Z}$

Thus the sum of all probabilities add to unity. The ratio of excited particles to particles in ground state is

$R=P_2/P_1=\frac{4\exp(-E_2/kT)}{\exp(-E_1/kT)}=4\exp[-(E_2-E_1)/(kT)]=4\exp (-\frac{(-3.4+13.6)*1.6*10^{-19}}{1.38*10^{-23}*10^7})=$

$=4*0.9882=3.953$

For the third excited state of Hydrogen ($n=3, l=2, m=2$) please find the expectation values of

a) $<r^2>$

b) $<x^2>$

For $n=3, l=2$ and $m=2$ the wavefunction of the electron in Hydrogen atoms is

$\psi_{322}=R_{32}(r)Y_2^2(\theta,\phi)$ with $R_{32}(r)=\frac{2\sqrt{2}}{27\sqrt{5}}(Z/3a)^{3/2}*(Zr/a)^2*\exp(-Zr/3a)$

where $a$ is the Bohr radius. The element of volume in spherical coordinates is

$dV=r^2dr*\sin\theta d\theta*d\phi$.

The expectation value for $<r^2>$ is ($Z=1$):

$<r^2> =\int_V \psi_{322}^*r^2\psi_{322}*dV=\int_0^a r^2R_{32}^2r^2dr*\int_S (Y_2^2)^*Y_2^2d\Omega=\int_0^a r^4R_{32}^2dr$

$<r^2>=\frac{8}{3645}*\frac{1}{27a^7}*\int_0^a r^8\exp(-2r/3a)*dr=$

$=\frac{8}{98415a^7} (\frac{6200145a^9}{4}-12076233*e^-{2/3}*a^9/4)$

$<r^2>=4.9662*10^{-6}*a^2$.

To find the expectation value of $<x^2>$ one needs also to know the shape of the $Y_2^2$ functions

$Y_2^2=\sqrt{\frac{15}{32\pi}}\exp(2i\phi)*\sin^2(\theta)$.

Therefore when $x^2=r^2\sin^2\theta*\cos^2\phi$ one has

$<x^2>= \int_0^a r^2R_{32}^2*r^2 dr*(\frac{15}{32\pi}\int_0^\pi\sin^2\theta \sin^4\theta *\sin\theta d\theta *\int_0^{2\pi}\cos^2\phi d\phi)$

because $\exp(2i\phi)\exp(-2i\phi)=1$

One has

$\int_0^a \sin^7\theta d\theta =32/35$  and $\int_0^{2\pi}\cos^2\phi d\phi =\pi$  so that

$<x^2>=4.9662*10^{-6}a^2(\frac{15}{32\pi}\frac{32}{35})\pi=2.128*10^{-6}*a^2$