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Emma Healy 2J
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Postby Emma Healy 2J » Tue Nov 17, 2020 10:58 pm

If n=6, would l=5 be a valid answer? I know that l can be 0,...n-1 and that l=0 corresponds to s, l=1 corresponds to p, l=2 corresponds to d, l=3 corresponds to f. Based on this, l=5 would not correspond to s, p, d, or f. So would l=5 be valid because it fits for n=6-1=5?

Stuti Pradhan 2J
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Re: n=6,l=5

Postby Stuti Pradhan 2J » Tue Nov 17, 2020 11:02 pm

Yes, l=5 is a valid answer. There are other orbitals past the s, p, d, and f orbitals (for example there is a g orbital that exists, but there are no elements that have been found that can occupy this orbital). Any orbital that satisfies the 0, ..., n-1 requirement is a possible value for l.

Hope this helps!

Hailey Kang 2K
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Re: n=6,l=5

Postby Hailey Kang 2K » Tue Nov 17, 2020 11:38 pm


l=5 would be a valid answer. Your l value can be all the number leading up to n-1. so for example, if n=5, all your possible l values would be 0,1,2,3,4.

Samudrala_Vaishnavi 3A
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Re: n=6,l=5

Postby Samudrala_Vaishnavi 3A » Mon Dec 07, 2020 8:09 am

Theoretically, that would work considering that the l number is one less than the n number. l is always one less than the n number considering it is a subshell of it.

sophie esherick 3H
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Re: n=6,l=5

Postby sophie esherick 3H » Mon Dec 07, 2020 9:22 am

Yes, l=5 is a valid answer because there are other orbitals past the s, p, d, and f orbitals. As long as the angular momentum quantum number follows the l=0,,, n-1 requirement, it is possible.

Chenning Yang Dis3l
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Re: n=6,l=5

Postby Chenning Yang Dis3l » Tue Dec 15, 2020 10:40 am

Yes! l =<(n-1)

Hannah Chang 3K
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Re: n=6,l=5

Postby Hannah Chang 3K » Tue Dec 15, 2020 1:24 pm

l=5 is theoretically correct though we haven't found actual elements with h orbitals.

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