The sum of electrons present in all sub shells of an atom with $m_s$ value of $+\frac{1}{2}$ for $n=4$ and…
The sum of electrons present in all sub shells of an atom with $m_s$ value of $+\frac{1}{2}$ for $n=4$ and $m_s$ value of $-\frac{1}{2}$ for $\mathrm{n}=3$ is
$25$
$16$
$09$
$32$
Solution
No. of electrons in ' $n$ ' shell $=2 n^2$
$\therefore \mathrm{n}=4 ; 2 \mathrm{n}^2=32 \mathrm{e}$ and $\mathrm{n}=3 ; 2 \mathrm{n}^2=18 \mathrm{e}$
Half of the electrons in one shell has $\left(+\frac{1}{2}\right)$ spin and half of them has $\left(-\frac{1}{2}\right)$ spin.
$\therefore \quad$ Total no. of electrons $=\frac{32}{2}+\frac{18}{2}=25 \mathrm{e}$