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Chemistry 8 Online
OpenStudy (anonymous):

For the principal quantum number n=2, how many possible values are there for the angular momentum quantum number (l)?

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?

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.

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

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.

OpenStudy (anonymous):

Didnt catch that

OpenStudy (anonymous):

Thanks!

OpenStudy (abhisar):

Is that clear now?

OpenStudy (anonymous):

Yes

OpenStudy (abhisar):

ur welcome :)

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