Question

The correct order of decreasing energy for the electrons whose quantum numbers, $u$ and $I$ 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 $l=1$

• A. $A>C>B>D$
• B. $A>B>C>D$
• C. $C>A>D>B$
• D. $A>B>D>C$

Answer 1

Answer: (A)

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 $ 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. $A>C>B>D$