Question

The atomic number of potassium is $19$. Hence the number of electrons in N shell will be
(A) $9$
(B) $1$
(C) $8$
(D) $2$

Answer 1

In an atom, the nucleus is present in the center and the electrons are present in various shells around the nucleus. The shells are alphabetically represented as $K,L,M,N....$ and so on. These shells can be figured out by writing the electronic configuration of the atom.

Complete step by step answer:
There are a total four quantum numbers that are $n,l,ml$ and $ms$. Among which $n$ or principal quantum. Number determines the shells of atoms. The size and energy of the atom depends upon its principal quantum number.
AS the number of $n$ increases, so increases the number of shells. The shells are represented as:

Principal quantum number ($n$)	Shell
$1$	$K$
$2$	$L$
$3$	$M$
$4$	$N$

In the given question, to find out the number of electrons in $N$ shell, we firstly need to know the electronic configuration of the atom that is potassium.
Atomic number of potassium $= 19$
Its, electronic configuration $= 2,8,8,1$
$K\,L\,M\,N$
Thus, as $N$ is the fourth shell, it is having one electron in it.
Here, $N$ shell is the outermost or the valence shell. So, potassium has one valence electron. Therefore, potassium belongs to the group $1$ of periodic table.
Potassium has $2,8$ and $8$ electrons in the $K,L$ and $M$ respectively.

So, the correct answer is Option B.

Additional Information:
The $n$ also determines the number of subshells within the shell. The subshells $s,p,d,f$ are determined by azimuthal quantum number ($l$). The value of $l$ always ranges between $0$ to $n - 1$.
Azimuthal quantum number is also known as orbital or subsidiary quantum number defines the three-dimensional shape of the orbital.

Note:
The maximum number of electrons present in a shell ($n$) can be determined by the formula $2{n^2}$. For example, the number of electrons in the first shell is, $n = 1$, $2{n^2} = 2$.
For the second shell, $n = 2$, $2{n^2} = 8$. Hence, $8$ electrons in the second shell and so on.