(a) Show that the number of electron states in a subshell is \(4 \ell+2 .\) (b) By summing the number of states in each of the subshells, show that the number of states in a shell is \(2 n^{2} .\) [Hint: The sum of the first \(n\) odd integers, from 1 to \(2 n-1,\) is \(n^{2} .\) That comes from regrouping the sum in pairs, starting by adding the largest to the smallest: \(1+3+5+\dots+(2 n-5)+(2 n-3)+(2 n-1)\) \(=[1+(2 n-1)]+[3+(2 n-3)]+[5+(2 n-5)]+\cdots\) \(=2 n+2 n+2 n+\cdots=2 n \times \frac{n}{2}=n^{2}\)

In a subshell, there are \((2\ell+1)\) degenerate (i.e., having the same energy) orbitals. Here, \(\ell\) is the angular momentum quantum number which ranges from \(0\) to \((n-1)\), where \(n\) is the principal quantum number of the shell. Each orbital can have \(2\) electrons, one with spin quantum number \(+1/2\) and the other with spin quantum number \(-1/2\). Thus, the total number of electron states in a subshell is \(2 \times (2\ell+1) = 4 \ell+2\).

To find the number of electron states in a shell, we sum the number of states in each of the subshells. We have the following set of subshells in a shell of principal quantum number \(n\): \(\ell = 0,1,2,...,n-1\). For each subshell, the number of electron states is given by \(4\ell + 2\). The sum of electron states in all subshells of a shell is: \((4 \times 0 + 2) + (4 \times 1+ 2) + (4 \times 2 + 2) + \cdots + (4 \times (n-1) + 2)\). We can now factor out a \(2\) from each term to get a sum of odd integers from \(1\) to \(2n-1\) (as hinted in the question): \(2(1 + 3 + 5 + \cdots + (n-1)2)\). Using the hint provided in the question, we know that the sum of the first \(n\) odd integers is \(n^2\). Thus, the sum of electron states in all subshells of a shell is: \(2 \times n^2 = 2n^2\). So, the number of states in a shell is \(2n^2\).