Given, (A) n =5, ~m ₁=-1 (B) n =3, l =2, ~m ₁=-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) respectively are :

Step-by-step solution

For case (A) with n=5, m_l=-1 : The principal quantum number n=5 allows l from 0 to 4. Since m_l=-1 , we need l≥1. For each l value (1, 2, 3, 4), the orbital m_l=-1 exists, giving 4 orbitals. Each orbital holds 2 electrons (spin up and down), so maximum electrons = 8. For case (B) with n=3, l=2, m_l=-1 , m_s=+ 1 2 : These quantum numbers completely specify a single electron in a unique orbital with a specific spin, so maximum electrons = 1. Therefore, the answer is 8 and 1.