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
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
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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 \(n^2\).
Updated On: Jul 18, 2026
- 25
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The Correct Option is A
Solution and Explanation
Step 1: Use the formula for total orbitals in a shell.
For a shell with principal quantum number \(n\), the total number of orbitals is \[ n^2 \] Each orbital can contain two electrons, one with \[ m_s=+\frac{1}{2} \] and one with \[ m_s=-\frac{1}{2} \]
Step 2: Electrons with \(m_s=+\frac{1}{2}\) for \(n=4\).
For \(n=4\), \[ \text{Number of orbitals}=n^2=4^2=16 \] Hence, the number of electrons having \[ m_s=+\frac{1}{2} \] is \[ 16 \]
Step 3: Electrons with \(m_s=-\frac{1}{2}\) for \(n=3\).
For \(n=3\), \[ \text{Number of orbitals}=n^2=3^2=9 \] Hence, the number of electrons having \[ m_s=-\frac{1}{2} \] is \[ 9 \]
Step 4: Find the required sum.
Therefore, the required sum is \[ 16+9=25 \]
Step 5: Final conclusion.
Hence, \[ \boxed{25} \]
For a shell with principal quantum number \(n\), the total number of orbitals is \[ n^2 \] Each orbital can contain two electrons, one with \[ m_s=+\frac{1}{2} \] and one with \[ m_s=-\frac{1}{2} \]
Step 2: Electrons with \(m_s=+\frac{1}{2}\) for \(n=4\).
For \(n=4\), \[ \text{Number of orbitals}=n^2=4^2=16 \] Hence, the number of electrons having \[ m_s=+\frac{1}{2} \] is \[ 16 \]
Step 3: Electrons with \(m_s=-\frac{1}{2}\) for \(n=3\).
For \(n=3\), \[ \text{Number of orbitals}=n^2=3^2=9 \] Hence, the number of electrons having \[ m_s=-\frac{1}{2} \] is \[ 9 \]
Step 4: Find the required sum.
Therefore, the required sum is \[ 16+9=25 \]
Step 5: Final conclusion.
Hence, \[ \boxed{25} \]
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