Calculate the maximum total number of electrons in Gallium(Z=31), with spin +1/2 and which occupy orbitals having minimum one radial node

Dear Student

Gallium
Atomic number, Z = 31
Column = 13 ;    Row = 4

Electronic Configuration --> 1s² 2s² 2p⁶ 3s² 3p⁶ 4s^​​2  3d¹⁰ 4p¹   or      [Ar] ​4s^​​2 3d¹⁰ 4p¹

4p here, has a single electron which can either be in 4p_x , 4p_y or 4p_z orbital (i.e., 3) and it can either be spin up or spin down (i.e., 2).
Therefore, possible set of quantum numbers = 3 x 2 = 6

For 4p orbital -
n = 1,2, ....., N = 4
So, the principal quantum number, N = 4

l = 0,1,2....., n -1 = 1
Angular momentum quantum number = 1

m_l = {0,1, ..., $\pm$l} = {0, ​$\pm$1} 
So, magnetic quantum number will be {0, ​$\pm$1}
Here, we have, 2l + 1 = 2(1) + 1 = 3 total 4p orbitals


For 4p electron -
n and l quantum number are fixed.
However, m_l may vary as -1, 0 or +1.
Spin quantum number, m_s can be ​​$\pm$$\frac{1}{2}$

Therefore, the possible set of quantum numbers are -
(n, l, m_l, m_s) = (4, 1, -1, +​$\frac{1}{2}$)
(n, l, m_l, m_s) = ​(4, 1, 0, +​$\frac{1}{2}$)
(n, l, m_l, m_s) = ​(4, 1, +1, +​$\frac{1}{2}$)
(n, l, m_l, m_s) = ​(4, 1, -1, -​$\frac{1}{2}$)
(n, l, m_l, m_s) = ​(4, 1, 0, -​$\frac{1}{2}$)
(n, l, m_l, m_s) = ​(4, 1, +1, -​$\frac{1}{2}$)

Radial Nodes -
For a 4p orbital, l = 1
Radial Node = n - l - 1 = 4 - 1-1 = 2
So, if there is a minimum of one radial node than one electron will occupy the same.


Regards