Is energy emitted or absorbed when the following electronic transitions occur in hydrogen? (a) from \(n=4\) to \(n=2\) , (b) from an orbit of radius 2.12 A to one of radius \(8.46 \hat{A},(\mathbf{c})\) an electron adds to the \(\mathrm{H}^{+}\) ion and ends up in the \(n=3\) shell?

Here, we need to determine if energy is emitted or absorbed during a transition from n=4 to n=2. First, let's observe their energy levels using the relation mentioned above: \(E_4 = -\cfrac{13.6 \text{ eV}}{4^2} = -0.85 \text{ eV}\) \(E_2 = -\cfrac{13.6 \text{ eV}}{2^2} = -3.40 \text{ eV}\) Since the electron transitions from a higher energy level (n=4) to a lower energy level (n=2), energy is emitted.

For this case, we have to find the energy levels corresponding to the given radii. The formula to find the radius, r, of the nth orbit in hydrogen is: \(r_n = 0.529n^2 \ Å\) Solving for n for the two given radii, we get: \(n_1^2 = \cfrac{2.12}{0.529} \Rightarrow n_1 = 2\) \(n_2^2 = \cfrac{8.46}{0.529} \Rightarrow n_2 = 4\) Here, an electron transitions from an orbit with a radius of 2.12 Å (n=2) to an orbit with a radius of 8.46 Å (n=4). Since the electron transitions from a lower energy level (n=2) to a higher energy level (n=4), energy is absorbed.

For this situation, we have an electron joining an H+ ion, which is a hydrogen atom without an electron. When the electron joins and ends up in the n=3 shell, it goes from free space, where its energy is approximately 0 eV, to the n=3 orbit in the hydrogen atom. We already know from the analysis that the energy level for n=3 is: \(E_3 = -\cfrac{13.6 \text{ eV}}{3^2} = -1.51 \text{ eV}\) Since the electron transitions from a higher energy level (0 eV) to a lower energy level (n=3), energy is emitted. In summary: (a) Energy is emitted during the transition from n=4 to n=2. (b) Energy is absorbed during the transition from 2.12 Å to 8.46 Å. (c) Energy is emitted when an electron adds to the H+ ion and ends up in the n=3 shell.