Answer
For a helium atom in its ground state, the quantum numbers $(n,l,m_l,m_s)$ of the spin-down electron is $(1,0,0,-\frac{1}{2})$
Work Step by Step
The helium atom has $2$ electrons. At ground state of the helium atom, $2$ electrons fit into the lowest-energy subshell, the $1s$ subshell with quantum numbers $(n,l,m_l,m_s)$ of $(1,0,0,-\frac{1}{2})$ and $(1,0,0,+\frac{1}{2})$ respectively. Thus, for a helium atom in its ground state, the quantum numbers $(n,l,m_l,m_s)$ of the spin-down electron is $(1,0,0,-\frac{1}{2})$
