The energy of the light emitted when a hydrogen electron goes from n = 2 to n = 1 is what fraction of its grou

Question

The energy of the light emitted when a hydrogen electron goes from n = 2 to n = 1 is what fraction of its ground-state ionization energy?

Answer 1

If the energy of K (n = 1) shell is E1 then the energy of the L (n = 2) shell equals:

E2 = E1/(2^2) = E1/4

So the energy of the light emitted when an electron jumps from L shell to K shell is:

ΔE = E1 - E2 = E1 - E1/4 = 3*E1/4

Now, the ground-state ionization energy Ei equals:

Ei = E1 - 0 = E1, so:

100*ΔE/E1 = 75%

Answer 2

The energy of a bound state of hydrogen is given by R/n^2, where R is the Rydberg constant (the size of which is irrelevant to this problem). All you have to do is to evaluate the energy for n = 1, n = 2, and n = infinity.