What is the magnetic quantum number of the 19th electron of Calcium?

##0##

of Calcium ##"Ca"## is 20.

We know that there are four 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 " to " (n-1)##; ##0=s, 1=p, 2=d, 3=f##

##m(l) =## magnetic quantum number indicating orbital within a sublevel. Range from ##-l" through "##0##" to "+l##

##m(s) =## spin quantum number. It identifies an electron within an orbital and can have either of the two values ##+1/2 and -1/2##

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The electronic configuration of calcium is: ##1"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 ##19th## electron goes to ##4"s"## energy level. As such for it ##n=4, l=0## We know that for ##l=0, m(l)=0## only. Hence magnetic quantum number ##m(l)## of ##19th## electron of calcium is ##=0##