How many unique combinations of the quantum numbers \(l\) and \(m_{l}\) are there when (a) \(n=1,(\mathbf{b}) n=5 ?\)

For (a) when n = 1, l ranges from 0 to (n-1), which is 0. For (b) when n = 5, l ranges from 0 to (n-1), which is 0, 1, 2, 3, 4.

(a) For n = 1, there is only one possible l value, which is 0. In this case, m_l can only be 0, giving us just one unique combination: l = 0 and m_l = 0. (b) For n = 5, there are five possible l values: 0, 1, 2, 3, 4. For each of these l values, we will calculate the unique combinations for m_l. - For l = 0, there is only one possible value for m_l, and that is 0. So, for l = 0, we have one unique combination: m_l = 0. - For l = 1, m_l can range from -1 to 1, giving us three unique combinations: m_l = -1, 0, 1. - For l = 2, m_l can range from -2 to 2, giving us five unique combinations: m_l = -2, -1, 0, 1, 2. - For l = 3, m_l can range from -3 to 3, giving us seven unique combinations: m_l = -3, -2, -1, 0, 1, 2, 3. - For l = 4, m_l can range from -4 to 4, giving us nine unique combinations: m_l = -4, -3, -2, -1, 0, 1, 2, 3, 4.

Now we need to calculate the sum of all the unique combinations for each l value when n = 5. Total unique combinations for n = 5 = 1 (for l=0) + 3 (for l=1) + 5 (for l=2) + 7 (for l=3) + 9 (for l=4) = 25. So, when n = 1, the total number of unique combinations for l and m_l is 1, and when n = 5, the total number of unique combinations for l and m_l is 25.