Angular Momentum of a hydrogen atom in the 7f state

AI Thread Summary
The discussion focuses on calculating the orbital angular momentum of a hydrogen atom in the 7f state using the formula L=sqrt(L(L+1)hbar). Initial calculations yielded an incorrect value of 1.41 hbar, which was later corrected to sqrt(12) or approximately 3.46 hbar. Participants emphasized the importance of maintaining accuracy to three significant figures throughout the calculations. The final values discussed were around 3.56 or 3.6 x 10^-34 J*s, indicating a need for precision in the final answer. The conversation highlights common pitfalls in quantum mechanics calculations and the importance of careful mathematical handling.
Noreturn
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Homework Statement


A hydrogen atom is in the 7f state.
What is the magnitude of its orbital angular momentum?

Homework Equations



L=sqrt(L(L+1)hbar

The Attempt at a Solution


L= Sqrt(3(3+1)Hbar)
1.41hbar (we want J*S)

1.41*1.054*10^-34

1.47*10^-34J*S
 
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Noreturn said:
1
2. Homework Equations

L=sqrt(L(L+1)hbar
Note that you have two left parentheses but only one right parenthesis.
3. The Attempt at a Solution
L= Sqrt(3(3+1)Hbar)
1.41hbar (we want J*S)
Did you want the hbar to be inside the square root?
I don't see how you got the 1.41 from the previous line. What number are you taking the square root of?
 
No sorry, so the anwser I got was 1.41 hbar.

So I ended up getting sqrt(3(4))hbar

sqrt(12)
or 3.46 h bar

So sorry guess 1.41 was wrong. However even with 3.56 or 3.6*10^-34 J*s was still wrong
 
Your work looks correct, now. But maybe you need to be accurate to three significant figures in your answer. Don't round off until the end.
 
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