Problem 8

  The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional 2.86 eV of energy. What is the quantum number n of the state into which the electron moves?

SOLUTION:
The energy levels of the hydrogen atom are given by (eqn. 30.13)

$$E_n = -(13.6 eV) \frac{1}{n^2}$$

In the first excited state, n = 2, so E₂ = -3.40 eV. It acquires 2.86 eV, so

E_n = -3.40 eV + 2.86 eV = -0.54 eV

$$-0.54 eV = -(13.6 eV)\frac{1}{n^2}$$

n = 5