For a helium atom in its ground state, what are quantum numbers $\left(n,l,m_I,m_s\right)$ for the (a) spin-up electron and (b) spin-down electron?

The helium atom is its ground state.

The set of numbers used to describe the position and power of an electron atom is called quantum numbers. There are four quantum numbers, namely, prime numbers, azimuthal, magnetic numbers, and spin quantum. The stored values of the quantum system are given by quantum numbers.

For every value of the principal quantum number, there are n values of l ranging from 0 to $\left(\mathrm{n}-1\right)$. For every value of the orbital quantum number, there exist $\left(2\mathrm{l}+1\right)$values of magnetic quantum number ranging from $-\mathrm{l}$to $+\mathrm{l}$. Thus, the maximum value of this magnetic quantum number is equal to the value of the orbital quantum number. Now, every electron has two spin orientations, thus the value of the magnetic quantum number is determined by these spins.

When the helium atom is in the ground state, both its electrons are in 1s state. Thus for each electron with quantum numbers n = 1 , l = 0 and ${\mathrm{m}}_{\mathrm{l}}=0$; one of the electrons is in a spin-up quantum state $\left({\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}\right)$and the other electron is in a spin-down $\left({\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}\right)$quantum state.

Hence, the quantum numbers for the spin-up electron are $\left(1,0,0+\frac{1}{2}\right)$.

Using the concept of calculations in above part (a), the quantum numbers for the spin-down electron is $\left(1,0,0-\frac{1}{2}\right)$.