Answer
The lithium atom rises to its first exited state, when the electron from $2s$ subshell goes to the $2p$ subshell. The electronic configuration of the atom at its first excited state becomes $1s^22p^1$. Thus the possible quantum numbers $(n,l,m_l,m_s)$ for the third electron in the first excited state are $(2,1,-1,\mp\frac{1}{2})$, $(2,1,0,\mp\frac{1}{2})$ and $(2,1,+1,\mp\frac{1}{2})$
Work Step by Step
The lithium atom has $3$ electrons. At ground state of the lithium atom, $2$ of $3$ 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. The remaining $1$ electron fill the next lowest energy subshell, the $2s$ subshell with quantum numbers $(n,l,m_l,m_s)$ of $(2,0,0,\mp\frac{1}{2})$
The lithium atom rises to its first exited state, when the electron from $2s$ subshell goes to the $2p$ subshell. The electronic configuration of the atom at its first excited state becomes $1s^22p^1$. Thus the possible quantum numbers $(n,l,m_l,m_s)$ for the third electron in the first excited state are $(2,1,-1,\mp\frac{1}{2})$, $(2,1,0,\mp\frac{1}{2})$ and $(2,1,+1,\mp\frac{1}{2})$
