Question:
The angular momentum of an electron in an orbit \(X\) of hydrogen atom is \(\frac{2h}{\pi}\). Maximum number of orbitals possible in \(X\) is
The angular momentum of an electron in an orbit \(X\) of hydrogen atom is \(\frac{2h}{\pi}\). Maximum number of orbitals possible in \(X\) is
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Maximum number of orbitals in a shell with principal quantum number \(n\) is \(n^2\).
Updated On: May 7, 2026
- \(4\)
- \(9\)
- \(16\)
- \(25\)
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The Correct Option is C
Solution and Explanation
Concept:
According to Bohr's model, angular momentum of an electron is:
\[
mvr=\frac{nh}{2\pi}
\]
where \(n\) is the principal quantum number.
Step 1: Given angular momentum: \[ \frac{nh}{2\pi}=\frac{2h}{\pi} \]
Step 2: Cancel \(h\) from both sides. \[ \frac{n}{2\pi}=\frac{2}{\pi} \]
Step 3: Multiply both sides by \(2\pi\). \[ n=4 \]
Step 4: Maximum number of orbitals in a shell is: \[ n^2 \] \[ n^2=4^2=16 \] Therefore, \[ \boxed{16} \]
Step 1: Given angular momentum: \[ \frac{nh}{2\pi}=\frac{2h}{\pi} \]
Step 2: Cancel \(h\) from both sides. \[ \frac{n}{2\pi}=\frac{2}{\pi} \]
Step 3: Multiply both sides by \(2\pi\). \[ n=4 \]
Step 4: Maximum number of orbitals in a shell is: \[ n^2 \] \[ n^2=4^2=16 \] Therefore, \[ \boxed{16} \]
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