Show that the number of different electron states possible for a given value of \(n\) is \(2 n^{2}\).

There are four quantum numbers for an electron in an atom: 1. Principle quantum number (\(n\)): defines the energy level or electron shell of the atom (\(n = 1, 2, 3, \ldots\)). 2. Angular momentum quantum number (\(\ell\)): describes the shape of the electron's orbitals (\(\ell = 0, 1, 2, \ldots, n - 1\)). 3. Magnetic quantum number (\(m_{\ell}\)): describes the orientation of the orbital in space (\(m_\ell = -\ell, -( \ell -1), \ldots, 0, \ldots, (\ell -1), \ell \)). 4. Spin quantum number (\(m_s\)): describes the electron's intrinsic angular momentum (either spin up or spin down, \(m_s = \pm \frac{1}{2}\)).

For a given value of \(n\), we have possible \(\ell\) values ranging from \(0\) to \(n - 1\). We can sum up the number of possible \(\ell\) values using the following formula: $$\sum_{\ell=0}^{n-1} (\ell+1)$$

Calculate the sum of the possible \(\ell\) values. $$\sum_{\ell=0}^{n-1} (\ell+1) = 1 + 2 + 3 + \cdots + n$$ This is the sum of an arithmetic series, which can be calculated using the formula: $$S_n = \frac{n(n+1)}{2}$$ In this case, \(S_n = \frac{n(n+1)}{2}\) is the sum of possible \(\ell\) values.

For each value of \(\ell\), there are \(2\ell + 1\) different unique values of \(m_\ell\). Also, for each value of \(m_\ell\), there are two possible values of \(m_s\). So the total number of possible electron states is given by: $$\text{Total electron states} = \sum_{\ell = 0}^{n-1} (2\ell + 1) \times 2$$

Now, substitute the sum formula from step 3 into the expression of the total electron states and simplify it. $$\text{Total electron states} = 2\sum_{\ell=0}^{n-1} (\ell + 1) = 2 \times \frac{n(n+1)}{2} = n(n+1)$$ For the given value of \(n\), we can see that the total number of electron states is equal to \(n(n+1)\).

Note that the exercise asks to prove that the number of electron states is equal to \(2n^2\). We can show that our result is equivalent to this target formula. For large values of \(n\), the term \(n + 1\) is approximately equal to \(n\) itself. So, the total number of electron states can be approximately written as: $$\text{Total electron states} = n(n+1) \approx n \times n = n^2$$ However, we must account for the two possible values of the spin quantum number \(m_s\), which was included in our calculations. Therefore, the final result is: $$\text{Total electron states} \approx 2 \times n^2$$ Thus, we have shown that for a given value of \(n\), the number of different electron states possible is approximately equal to \(2n^2\).