The maximum number of electrons which have n = 3 and m_s = +1/2 is

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answer is D.

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Detailed Solution

When n=3,l=0,1,2 i.e., there are 3s, 3p and 3d orbitals.
Total 18 electrons, 9 electrons with s=+½ and 9 with s=-½
Alternatively, in any n_^thorbit, there can be a maximum of 2n² electrons.

Hence, when n=3, number of maximum electrons =18. Out of these 18 electrons, 9 can have spin -½ and remaining nine with spin =+½

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