Question

How many electrons in an atom may have the following quantum numbers?
(a) n = 4, $m=-\frac{1}{2}$
(b) n = 3, l = 0

Answer 1

(a) Total number of electrons in an atom for a value of n = 2$n^2$
$\therefore$ For n = 4,
Total number of electrons = 2 $(4{)}^2=32$
The given element has a fully filled orbital as
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}$.
Hence, all the electrons are paired.
$\therefore$ Number of electrons (having n = 4 and $m_{-\frac{1}{2}}=(32\times \frac{1}{2})$ =16 electrons

(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.