Orbital angular momentum possible values of an electron in a hydrogen atom

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The possible values for the orbital angular momentum of an electron in a hydrogen atom with a principal quantum number n = 3 are determined by the quantum number l, which can take values of 0, 1, and 2. Using the formula L = √(l(l+1))ħ, the corresponding angular momentum values are 0, √2ħ, and √6ħ. The calculations confirm that these values are correct. The discussion concludes with a confirmation of the solution's accuracy.
Ezequiel
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Homework Statement



What are the orbital angular momentum possible values for an electron in a hydrogen atom with a principal quantum number n = 3?

Homework Equations



L = \sqrt{l(l+1)}\hbar

The Attempt at a Solution



Possible values for l are 0, 1, 2.

So, substituting these in the equation above I get: 0, √2\hbar, √6\hbar.

Is this the right answer?
 
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yep, that looks right.
 
Thank you again :)
 

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