The correct order of decreasing energy for the electrons whose quantum numbers $n$ and $l$ are given below, is A. $n=5$ and $l=2$ B. $n=5$ and $l=0$ C. $n=4$ and $l=3$ D. $n=4$ and $l=1$

Given, $n=$ principal quantum number The principal quantum number is one of four quantum numbers assigned to each electron in an atom to describe that electron state. As $n$ increases, the electron is also at a higher energy and is, therefore less tightly bound to the nucleus. $l=$ azimuthal quantum number The azimuthal quantum number is a quantum number for an atomic orbital that determines, it orbital angular momentum and describes the shape of the orbital. Azimutal quantum number $(l)=0,1,2,3,4$ Shape of orbitals $=s, p, d, f, g$ Energy of orbitals $s < p < d < f < g$ So, higher the value of $n$ and $l$ higher the energy of electron. Decreasing order of energy for the electrons, $n=5$ and $l=2>n=4$ and $l=3>n=5$ and $l=0>n=4$ and $l=1$ i.e. $\quad A>C>B>D$