Question

An excited hydrogen atom emits a photon with an energy of \(1.133 \mathrm{eV}\). What were the initial and final states of the hydrogen atom before and after emitting the photon?

Short answer

Question: In a hydrogen atom, a photon with an energy of 1.133 eV is emitted, causing a transition in the electron's energy level. Identify the initial and final energy levels of the hydrogen atom involved in this transition. Answer: The initial state of the hydrogen atom was in the 3rd energy level, and after emitting the 1.133 eV photon, the final state was in the 2nd energy level.

Step-by-step solution

Write the energy difference equation

We need to find the energies \(E_{n_1}\) and \(E_{n_2}\), such that their difference equals the energy of the emitted photon: \(E_{n_1} - E_{n_2} = 1.133 \mathrm{eV}\).

Express energies in terms of n

Using the formula \(E_n = -\dfrac{13.6 \mathrm{eV}}{n^2}\), write the energies \(E_{n_1}\) and \(E_{n_2}\) in terms of \(n_1\) and \(n_2\): \(-\dfrac{13.6 \mathrm{eV}}{n_1^2} - \left(-\dfrac{13.6 \mathrm{eV}}{n_2^2}\right) = 1.133 \mathrm{eV}\).

Simplify the equation

Combine the terms in the equation: \(\dfrac{13.6 \mathrm{eV}}{n_2^2} - \dfrac{13.6 \mathrm{eV}}{n_1^2} = 1.133 \mathrm{eV}\).

Find suitable values for n1 and n2

It can be time-consuming to find the exact values of \(n_1\) and \(n_2\) by trial and error. However, we can use the fact that energy differences between energy levels typically decrease as n increases. From this, it is reasonable to assume that the energy difference corresponds to a transition between low energy levels, likely from n = 2 or n = 3. Try combinations of n = 2, 3, or 4 and see which one gives a difference of about \(1.133 \mathrm{eV}\). After testing these values, we find that: \(\dfrac{13.6 \mathrm{eV}}{2^2} - \dfrac{13.6 \mathrm{eV}}{3^2} \approx 1.133 \mathrm{eV}\). Therefore, the initial and final states of the hydrogen atom are \(n_1 = 3\) and \(n_2 = 2\). In conclusion, the initial state of the hydrogen atom was in the 3rd energy level and after emitting the \(1.133 \mathrm{eV}\) photon, the final state was in the 2nd energy level.