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?

Answer 1

It seems like I'm answering this 5 years later, but this is for anyone who needs this in the future!

Helium is a hydrogen-like 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.