Question

For the $19^{th}$ electron of $K$ the values of quantum number will be

• A. $4$ , $1$ , $0$ , $+\frac{1}{2}$
• B. $4$ , $0$ , $0$ , $+\frac{1}{2}$
• C. $3$ , $2$ , $0$ , $+\frac{1}{2}$
• D. $3$ , $0$ , $0$ , $+\frac{1}{2}$

Answer 1

Answer: (B)

The electronic configuration of potassium, $K$ is
${}_{19}K=[Ar]4s^{-1}$
The $19th$ electron enters in $4s$-orbital.
For $4s^{-1}$
$n=4$
$l=0$ $(\because$ for $s$ orbital, $l=0)$
$m=0$
$s=+\frac{1}{2}$
Thus, the values of quantum number for $19th$ electron of $K$ will be $4,0,0,1/2$