The number of electrons that can be present i | vedclass

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(C) For $n=1$,the orbital is $1s$ ($1$ orbital). For $n=2$,the orbitals are $2s$ and $2p$ ($1+3=4$ orbitals). For $n=3$,the orbitals are $3s$,$3p$,and

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(C) For $n=1$,the orbital is $1s$ ($1$ orbital). For $n=2$,the orbitals are $2s$ and $2p$ ($1+3=4$ orbitals). For $n=3$,the orbitals are $3s$,$3p$,and $3d$ ($1+3+5=9$ orbitals). The total number of orbitals for $n$ up to $3$ is $1 + 4 + 9 = 14$. Each orbital can accommodate exactly one electron with a spin quantum number $m_s = -\frac{1}{2}$. Therefore,the total number of electrons with $m_s = -\frac{1}{2}$ is $14$. Hence,the correct option is $C$.

The number of electrons that can be present in sub-shells having $m_s$ value of $\frac{-1}{2}$,for $n$ up to $3$.

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