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 $4, 0, 0, 1/2$