Posted by qwer on .
In the Sun, an ionized helium atom makes a transition from n =6 state to the n = 2 state, emitting a photon. Can that photon be absorbed by hydrogen atoms present in the Sun? If so between what energy states will the hydrogen atom jump?

Modern Physics 
Omar,
It seems like I'm answering this 5 years later, but this is for anyone who needs this in the future!
Helium is a hydrogenlike atom. Using the fact that the ground state energy of hydrogen is 13.6eV, we can use the relation:
E = 13.6eV(z^2/n^2) to calculate the energy levels of helium, where z is charge number and n^2 is the principle quantum number.
For Helium z = 2, and in our case n1 = 6 and n2 = 2.
Therefore,
Change in E = 13.6eV*z^2* (1/(n2^2)  1/(n1^2))
That is,
Change in E = 13.6*4*((1/4)(1/36)) = 12.1eV.
In words, the ionised helium atom's photon emission requires 12.1eV, and that is the energy of the photon.
Now, this 12.1eV must be compared with hydrogen state transition energies.
For H:
Energy(in a given state) = 13.6eV(1/(n^2))
I will replace n by m here to avoid confusion, m for hydrogen where n was used for helium.
By comparing changes in energy between levels, where m1 = 1 and m2 = 3 yields a change in energy = 12.1eV.
So a hydrogen atom will absorb the photon and experience a transition, as above.