Given:(A) n = 5, m_l = +1(B) n = 2, l = 1, m_ | vedclass

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(B) $(A)\ n = 5$ implies $l$ can be $0, 1, 2, 3, 4$ ($s, p, d, f, g$ orbitals).For a given $m_l = +1$,the orbitals that can have this value are $p$ $(

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$8$ and $1$

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(B) $(A)\ n = 5$ implies $l$ can be $0, 1, 2, 3, 4$ ($s, p, d, f, g$ orbitals).For a given $m_l = +1$,the orbitals that can have this value are $p$ $(l=1)$,$d$ $(l=2)$,$f$ $(l=3)$,and $g$ $(l=4)$.Each orbital can hold $2$ electrons $(m_s = +1/2, -1/2)$.Total electrons $= 2 (p) + 2 (d) + 2 (f) + 2 (g) = 8$ electrons.$(B)\ n = 2, l = 1, m_l = -1, m_s = -1/2$ represents a specific electron in the $2p$ orbital.Since all four quantum numbers are specified,it corresponds to exactly $1$ electron.

Given:
$(A)\ n = 5, m_l = +1$
$(B)\ n = 2, l = 1, m_l = -1, m_s = -1/2$
The maximum number of electron$(s)$ in an atom that can have the quantum numbers as given in $(A)$ and $(B)$ are respectively:

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Given:$(A)\ n = 5, m_l = +1$$(B)\ n = 2, l = 1, m_l = -1, m_s = -1/2$The maximum number of electron$(s)$ in an atom that can have the quantum numbers as given in $(A)$ and $(B)$ are respectively: