Question.
How many electrons in an atom may have the following quantum numbers?
(a) $n=4$
$m_{x}=-\frac{1}{2}$
(b) $n=3, I=0$
How many electrons in an atom may have the following quantum numbers?
(a) $n=4$
$m_{x}=-\frac{1}{2}$
(b) $n=3, I=0$
Solution:
(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
$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10}$
Hence, all the electrons are paired.
$\therefore$ Number of electrons (having $n=4$ and $\left.m_{x}=-\frac{1}{2} \quad\right)=16$
(b) $n=3, I=0$ indicates that the electrons are present in the $3 s$ orbital. Therefore,
the number of electrons having $n=3$ and $I=0$ is 2 .
(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
$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10}$
Hence, all the electrons are paired.
$\therefore$ Number of electrons (having $n=4$ and $\left.m_{x}=-\frac{1}{2} \quad\right)=16$
(b) $n=3, I=0$ indicates that the electrons are present in the $3 s$ orbital. Therefore,
the number of electrons having $n=3$ and $I=0$ is 2 .