Chemistry
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OpenStudy (anonymous):
For the principal quantum number n=2, how many possible values are there for the angular momentum quantum number (l)?
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OpenStudy (anonymous):
1
2
4
6
OpenStudy (anonymous):
@Abhisar
OpenStudy (anonymous):
@mathslover @iGreen
OpenStudy (abhisar):
What do you think it should be?
OpenStudy (anonymous):
2?
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OpenStudy (abhisar):
How?
OpenStudy (anonymous):
Not really sure
OpenStudy (anonymous):
principal quantum number n=2
OpenStudy (abhisar):
For any value of n, possible values of l can be 0 to n-1
OpenStudy (abhisar):
Say, if n=3, then possible values of l will be 0,1 & 2 i.e there will be 3 posssible values of l.
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OpenStudy (anonymous):
so its 1 then?
OpenStudy (abhisar):
So, what will be the possible values of l for n=2?
OpenStudy (anonymous):
0, 1
OpenStudy (abhisar):
Yes :)
OpenStudy (anonymous):
thanks
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OpenStudy (abhisar):
So the correct option will be?
OpenStudy (anonymous):
A
OpenStudy (abhisar):
Nerps....you need to select the NUMBER of possible values and not the possible value..getting it..?
OpenStudy (anonymous):
oh okay. so it would be b then
OpenStudy (abhisar):
Yes, 0 & 1 are the possible values which means that there are 2 possible values.
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OpenStudy (anonymous):
Didnt catch that
OpenStudy (anonymous):
Thanks!
OpenStudy (abhisar):
Is that clear now?
OpenStudy (anonymous):
Yes
OpenStudy (abhisar):
ur welcome :)