# Question 29f4d

Jun 12, 2016

$0$

#### Explanation:

Atomic number of Calcium $\text{Ca}$ is 20.

We know that there are four quantum numbers which can be assigned to "describe" any electron in an atom. Each electron has a unique set of these four quantum numbers.

$n =$ principal quantum number, which defines the energy level of the electron.

$l =$ azimuthal (angular) quantum number defining the energy sub-level. Its values range from $0 \text{ to } \left(n - 1\right)$; $0 = s , 1 = p , 2 = d , 3 = f$

$m \left(l\right) =$ magnetic quantum number indicating orbital within a sublevel. Range from $- l \text{ through }$0$\text{ to } + l$

$m \left(s\right) =$ spin quantum number. It identifies an electron within an orbital and can have either of the two values $+ \frac{1}{2} \mathmr{and} - \frac{1}{2}$

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The electronic configuration of calcium is:
$1 {\text{s"^2; 2"s"^2, 2"p"^6; 3"s"^2, 3"p"^6; 4"s}}^{2}$ or ["Ar"] 4"s"^2#

it is evident that the $19 t h$ electron goes to $4 \text{s}$ energy level.
As such for it $n = 4 , l = 0$
We know that for $l = 0 , m \left(l\right) = 0$ only.
Hence magnetic quantum number $m \left(l\right)$ of $19 t h$ electron of calcium is $= 0$