# Question #a14c2

Jul 20, 2017

$9 \cdot {a}_{0}$

#### Explanation:

All you need to know in order to answer this question is that the radius of an orbit depends on the principal quantum number, $n$, that describes the energy level on which the electron is located.

If you take ${a}_{0}$ to be the Bohr orbit, you can say that you have

${r}_{n} = {n}^{2} \cdot {a}_{0}$

In your case, the third orbit is characterized by

$n = 3$

so you will have

${r}_{3} = {3}^{2} \cdot {a}_{0} = 9 \cdot {a}_{0}$