Question

For the quantum number lower case L = 0, how many possible values are there for the quantum number m subscrip l?

Answer 1

Well first off, l=0, l=1, l=2, and l=3 are all exactly the same as s, p, d, and f orbitals. There's only one s orbital that can hold a pair of electrons at a time, so m=0 is the only possible value you can have. However you probably already know there are 3 p orbitals, each of which can also hold a pair of electrons. Those 3 p orbitals are really represented by the m=-1, m=0, and m=1. Similarly we also have 5 d orbitals represented by m=-2,-1,0,1,2.

Fairly straight forward I think since s orbital is l=0 we have m=0. Since p orbital is l=1 we have m=-1,0,1. So all the m values are really just the range from negative to positive of the l value, not so bad.

Answer 2

The answer choices are... 0,1,2,3

Answer 3

Assuming this is the hydrogen atom quantum numbers, note the orbital quantum number l has \[l = 0, 1,2,3,4...(n-1)\] and the magnetic quantum number which is \[m_l = 0, \pm 1, \pm 2, \pm 3,...\pm l\] and just for some extra information the principal quantum number \[n = 1, 2, 3, 4, ... \infty \]

Answer 4

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