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Q.
The maximum number of electrons, present in an orbit that is represented by azimuthal quantum number $(l)=3$, will be
AIPMTAIPMT 1991Structure of Atom
Solution:
When azimuthal quantum number is $3$
$m =(21+1) $
$1 =3 $
$m =(2 \times 3+1)$
$=7 $ orbitals
then total values of $m=(2 \times 3+1)=7$ orbitals.
We know that, one orbital contains two electrons.
Hence, total number of electrons $=7 \times 2=14$.
Alternative
Total number of electrons $=4 l+2$
$=4 \times 3+2=12+2=14 $ electrons