Question

For a hydrogen like ion $H{e}^{+}$, there is a dumb-belled symmetric state A with 2 radial nodes in excited state. The final state after de-excitation was found to be ground state of H - atom.
How many electrons with spin quantum number $m_s=-\frac{1}{2}$ can be placed in A?

• A. 3
• B. 10
• C. 2
• D. 14

Answer 1

The correct option is A 3

State A has l = 1 and n = 4
If means it’s a 4p subshell.
Hence maximum number of electrons possible = 6, electrons with $m_s=-\frac{1}{2}$ is 3.