Arrange the following orbitals in decreasing order of energy. the order is:
$\mathrm{A}. \mathrm{n}=3,\mathrm{l}=0,\mathrm{m}=0$
$\begin{array}{r}\mathrm{B}. \mathrm{n}=4,\mathrm{l}=0,\mathrm{m}=0\\ \mathrm{C}.\mathrm{n}=3,\mathrm{l}=1,\mathrm{m}=0\\ \mathrm{D}.\mathrm{n}=3,\mathrm{l}=2,\mathrm{m}=1\end{array}$
The correct option for the order is

$\text{ Energy of orbitals }\mathrm{\alpha }(\mathrm{n}+\mathrm{l})$
If n + l value is same than orbital with higher ‘n ’ value have higher energy
$\begin{array}{r}\mathrm{A}.\mathrm{n}=3,\mathrm{l}=0,\mathrm{m}=0\Rightarrow \mathrm{n}+\mathrm{l}=3\\ \mathrm{B}.\mathrm{n}=4,\mathrm{l}=0,\mathrm{m}=0\Rightarrow \mathrm{n}+\mathrm{l}=4\\ \mathrm{C}.\mathrm{n}=3,\mathrm{l}=1,\mathrm{m}=0\Rightarrow \mathrm{n}+\mathrm{l}=4\\ \mathrm{D}.\mathrm{n}=3,\mathrm{l}=2,\mathrm{m}=1\Rightarrow \mathrm{n}+\mathrm{l}=5\end{array}$
Energy: $\mathrm{D}>\mathrm{B}>\mathrm{C}>\mathrm{A}$