The electrons, identified by quantum numbers $$n$$ and $$l,$$
(i) $$n = 4,\,l = 1,$$   (ii) $$n = 4,\,l = 0,$$   (iii) $$n = 3,\,l = 2,$$   and (iv) $$n = 3,\,l = 1,$$   can be placed in order of increasing energy, from the lowest to highest, as

Solution :
TIPS/Formulae : The two guiding rules to arrange the various orbitals in the increasing energy are:
(i) Energy of an orbital increases with increase in the value of $$n + l.$$
(ii) Of orbitals having the same value of $$n + l,$$ the orbital with lower value of n has lower energy.
Thus for the given orbitals, we have
$$\eqalign{ & ({\text{i}})\,\,n + l = 4 + 1 = 5 \cr & {\text{(ii)}}\,n + l = 4 + 0 = 4 \cr & {\text{(iii)}}\,n + l = 3 + 2 = 5 \cr & {\text{(iv)}}\,n + l = 3 + 1 = 4 \cr} $$
Hence, the order of increasing energy is (iv) < (ii) < (iii) < (i)