Problem: The orbital angular momentum of an electron has a magnitude of 4.716 × 10–34 kg•m2/s. What is the angular-momentum quantum number l for this electron?

FREE Expert Solution

The orbital angular momentum, L:

L=l(l+1), where ℏ = h/2π and h is the Plank's constant.

Solving for l:

L2=l(l+1)h24π2l(l+1)=L24π2h2l2+l=(4.716×10-34)2(4)(π)2(6.626×10-34)2

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The orbital angular momentum of an electron has a magnitude of 4.716 × 10–34 kg•m2/s. What is the angular-momentum quantum number l for this electron?

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