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The orbital angular momentum of \( 3^{\text {rd }} \) orbital is equal to \( \sqrt{\mathrm{x}} \frac{\mathrm{h}}{2 \pi} \) the value of ' \( \mathrm{x} \) ' is
ChemistryStructure of AtomJEE Main
Solution:
1046 Upvotes Verified Answer
The correct answer is: 6

Orbital angular momentum = l(l+1) h/2π,

Equate with the given orbital angular momentum value, calculate the value of x as:

2(3)h2π=6h2π.

So, x=6.

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