Question

Arrange the electrons represented by the following sets of quantum numbers in decreasing order of energy.
(i) $n=4,l=0,m_l=0,m_s=+\frac{1}{2}$
(ii) $n=3,l=1,m_l=1,m_s=-\frac{1}{2}$
(iii) $n=3,l=2,m_l=0,m_s=+\frac{1}{2}$
(iv) $n=3,l=0,m_l=0,m_s=-\frac{1}{2}$

Answer 1

The n with the largest value will have the highest energy. Then look at the value of the " $l$ ". The $l$ with the highest value will have the highest energy if the value of the "$n$" are equal.
So the lowest in energy will be (iv) because $l=0$
then the next lowest will be (ii) because $l=1$
then the next lowest will be (iii) because $l=2$
The highest in energy will be (i) since $n=4$