Question

The sum of electrons present in all subshells of an atom with \(m_s\) value of \(+\frac{1}{2}\) for \(n=4\) and \(m_s\) value of \(-\frac{1}{2}\) for \(n=3\) is

• A. 25
• B. 16
• C. 09
• D. 32

Hint:

For any shell with principal quantum number \(n\), the number of orbitals is \(n^2\). Therefore, the number of electrons with one particular spin value is also \(n^2\).

Answer 1

The Correct Option is A

Step 1: Recall electron capacity per subshell.
For a shell of principal quantum number $n$, the total electron capacity is $2n^2$, split equally between $m_s=+\frac{1}{2}$ and $m_s=-\frac{1}{2}$, so each spin type gets $n^2$ electrons when the shell is fully occupied.

Step 2: Count electrons with $m_s=+\frac{1}{2}$ in $n=4$.
A fully filled $n=4$ shell (4s, 4p, 4d, 4f) holds $2(4)^2=32$ electrons in total, so exactly half of them, $16$, carry $m_s=+\frac{1}{2}$.

Step 3: Count electrons with $m_s=-\frac{1}{2}$ in $n=3$.
A fully filled $n=3$ shell holds $2(3)^2=18$ electrons, so $9$ of them carry $m_s=-\frac{1}{2}$.

Step 4: Add the two counts.
\[ 16+9=25 \]
\[ \boxed{25} \]