What is the angular momentum of a hydrogen atom in different orbit states?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the angular momentum of a hydrogen atom in specific orbital states, namely the 4p and 5f states. Participants are exploring the relationship between quantum numbers and angular momentum in the context of atomic physics.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of quantum numbers to determine angular momentum, with some attempting calculations based on provided equations. Questions arise regarding the correct application of these quantum numbers, particularly the distinction between the principal quantum number 'n' and the orbital quantum number 'l'.

Discussion Status

Some participants have offered insights into the correct interpretation of quantum numbers, while others are verifying their calculations and seeking clarification on the definitions of 'l' in relation to the angular momentum formula. There is an ongoing exploration of the correct approach to the problem without a clear consensus yet.

Contextual Notes

Participants are working within the constraints of homework guidelines, which may limit the information they can use or share. The discussion reflects a mix of initial misunderstandings and subsequent clarifications regarding the quantum mechanical concepts involved.

Zywo
Messages
10
Reaction score
0

Homework Statement



What is the angular momentum of a hydrogen atom in (a) a 4p state and (b) a 5f state? Give your answers as a multiple of h-bar aka (h/2*pi)


Homework Equations



Radius of nth orbit = 5.29*10-11 * n^2
Angular Momentum = mvr
V of nth orbit = sqrt ( e^2 / (4*pi*epsilon_0*m*r) )

The Attempt at a Solution



For 4p orbit i plugged in 4 to the equations and got angular momentum is equal to 4 h-bar,
not right however...
Thanks,
David
 
Physics news on Phys.org
Solved my own problem, if anyone has this same issue the answer is more simple than i thought. L = h-bar*sqrt(l(l+1)) where l is the orbital quantum number
 
I tried plugging in 4 for the orbital quantum number and got it to be sqrt(20)*h-bar, but this was incorrect. Did I plug in the wrong number for the orbital?

Thanks,
Spencer
 
In the 4p state, n=4 and l=1. Where 'l' comes from the p in 4p. Remember the quantum number 'l' is labeled by s,p,d,f... for l=0,1,2,3,..

You don't use the 'n' when solving for the angular momentum.
 
Ok, you I forgot how to find l. Now it works. Thanks for the response
 

Similar threads

Replies
1
Views
2K
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
Replies
5
Views
2K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
Replies
3
Views
3K