Two of the three electrons in a lithium atom have quantum numbers $\left(n,I,m_I,m_s\right)$of $\left(1,0,0,+\frac{1}{2}\right)$and $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}$. What quantum numbers are possible for the third electron if the atom is (a) in the ground state and (b) in the first excited state?

Quantum numbers of two of the three electrons of the lithium atoms are $(1,0,0,+\frac{1}{2})$and $(1,0,0,-\frac{1}{2})$.

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.

The four quantum numbers represent the four quantum states of the individual electrons in a multi-electron state. Now, for the third electron of the lithium atom, the lowest energy level to be filled at n=2 , I=0. Again the next lowest state in energy is given by the state n=2 , I=1 , and other possible values of the magnetic quantum number.

The given two electrons have completely filled the lower state n=1 and thus the third electron now occupies the state n=2.

Now, from the above state n=2, the lowest energy level orbital quantum number is,I=0

, from possible values, I=0,1.

Again, the allowed value of the magnetic quantum number for I=0 is ${\mathrm{m}}_{\mathrm{I}}=0$.

But, every electron has two possibilities of spin quantum number that is $m_s=\pm \frac{1}{2}$.

Since there are no external magnetic fields, the energies of the two atoms are the same.

Hence, the possible cases of the quantum numbers of the third electron in the ground state are $\left(2,0,0,+\frac{1}{2}\right)$and $\left(2,0,0,-\frac{1}{2}\right)$.

The next energy level that is the first excited state, above the ground energy level has the state with the possible case of the lowest energy as:n=2, I=1 .

Again, the allowed values of the magnetic quantum number for I=1 are ${\mathrm{m}}_{\mathrm{I}}=-1,0,+1$.

But, every electron has two possibilities of spin quantum number that is $m_s=\pm \frac{1}{2}$.

Hence, the possible quantum numbers for the first excited state are $\left(2,1,1+\frac{1}{2}\right),\left(2,1,1-\frac{1}{2}\right),\left(2,1,0+\frac{1}{2}\right),\left(2,1,0-\frac{1}{2}\right),\left(2,1,-1+\frac{1}{2}\right)\mathrm{and}\left(2,1,-1-\frac{1}{2}\right)$.