Answer
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Work Step by Step
An ionized helium atom is just like hydrogen, except that there are two protons in the nucleus. Use equation 27–15b with Z=2.
$$E_{n}=-\frac{Z^2(13.6eV) }{n^2}=-\frac{2^2(13.6eV) }{n^2}=-\frac{54.4eV }{n^2}$$
Calculate the energy of a photon falling from the n = 6 state down to n = 2 state in singly ionized helium.
$$\Delta E=E_6-E_2=-(54.4eV)(\frac{1}{6^2}-\frac{1}{2^2})=12.1eV$$
Yes, the photon can be absorbed by a hydrogen atom in the sun. This is exactly the energy of a photon absorbed when an electron transitions from the n = 1 state up to the n = 3 state in hydrogen.
