The orbital angular momentum of a 4p electron will be — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Formula for Orbital Angular Momentum In wave mechanics, the orbital angular momentum ($L$) of an electron is quantized: $$L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi}$$ Where $l$ is the azimuthal quantum number. ### Step 2 - Quantum Numbers for a 4p Electron For a $4p$ electron: * Principal quantum number: $n = 4$ * Subshell: $p$ → azimuthal quantum number $l = 1$ *Note: The principal quantum number $n$ does NOT appear in the orbital angular momentum formula.* ### Step 3 - Calculate $L$ $$L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} = \sqrt{1(1 + 1)} \cdot \frac{h}{2\pi} = \sqrt{2} \cdot \frac{h}{2\pi}$$ $$L = \boxed{\sqrt{2} \cdot \frac{h}{2\pi}}$$ ### Step 4 - Evaluation of Options * **Option (A) $4 \cdot \frac{h}{2\pi}$:** This is Bohr's total angular momentum for the 4th orbit ($mvr = \frac{nh}{2\pi}$) — a different quantity. Incorrect. * **Option (B) $\sqrt{2} \cdot \frac{h}{2\pi}$:** Correct — matches our calculation. * **Option (C) $\sqrt{6} \cdot \frac{h}{4\pi}$:** Incorrect — denominator should be $2\pi$, not $4\pi$. * **Option (D) $\sqrt{2} \cdot \frac{h}{4\pi}$:** Incorrect — denominator should be $2\pi$, not $4\pi$. $$\text{Correct Option: } \boxed{\text{B}}$$