Question

In helium (${\text{1s}}^2$), if the spin quantum number of one electron is (${\text{m}}_{\text{s}}$) = -1/2, then the spin quantum number of the other electron would be ___.

• A. -1/2
• B. +1/2
• C. 0.0
• D. 1

Answer 1

The correct option is B +1/2

According to the Pauli's exclusion principle, the two electrons of the same atom cannot have the same four quantum numbers.

As an orbital can contain a maximum of only two electrons, the two electrons must have opposite spins or anti-parallel spins. This means if one is assigned an up-spin (+1/2), the other must be down-spin (-1/2).

In helium, both the electrons in the $1\mathrm{s\; orbital}$ have n = 1, l = 0, ${\text{m}}_{\text{l}}$= 0. If the spin quantum number of one electron is -1/2, then other would be +1/2 or vice-versa.