# Question #01331

Mar 23, 2017

3s

#### Explanation:

You can see the relation between Energy of the orbitals and principal quantum number as below
$E = - \left(\frac{13.6}{n} ^ 2\right) e V$

So, 2s orbital has $E = - \left(\frac{13.6}{2} ^ 2\right) e V = - 3.4 e V$
and
3s orbital has $E = - \left(\frac{13.6}{3} ^ 2\right) e V = - 1.51 e V$

So, 2s has lower energy and it is stable,
and 3s has higher energy.