Question

How many electrons in an atom may have the following quantum numbers?
(a) n = 4, m=−12 m=-\frac{1}{2} 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$n^2$n2 n^{2}

$\therefore$ For n = 4,
Total number of electrons = 2($4^2$) = 32
The given element has a fully filled orbital as
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}$
1s22s22p63s23p64s23d10 1s^{ 2} 2s^{ 2} 2p^{ 6} 3s^{ 2} 3p^{ 6} 4s^{ 2} 3d^{ 10} Hence, all the electrons are paired.
So there are total 32 electrons then there must be a 16 orbitals and each orbital contain 2 electrons one have m = -$\frac{1}{2}$ and other have m value =$\frac{1}{2}$ .
So there are 16 electrons that have quantum number n = 4 and m =$\frac{1}{2}$
(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.