To tackle this problem, we need to break it down into manageable parts. We have a sample of hydrogen gas with atoms in various excited states, and we want to find the principal quantum numbers of the initially excited electrons and the energies of the emitted photons. Let's go through this step by step.
Understanding Quantum States in Hydrogen
In a hydrogen atom, electrons occupy specific energy levels, which are defined by principal quantum numbers (n). The energy of an electron in a given state can be calculated using the formula:
E_n = -13.6 eV / n²
Here, E_n is the energy of the electron at the principal quantum number n. The negative sign indicates that the electron is bound to the nucleus. When an electron transitions from a higher energy level to a lower one, it emits a photon with energy equal to the difference between these two energy levels.
(a) Determining Principal Quantum Numbers
We know that the sample emits three different photons when irradiated with energy of 2.85 eV. The energy of the emitted photons must correspond to the energy differences between the excited states of the electrons. The maximum energy of the emitted photons is given as 13.4 eV.
To find the principal quantum numbers of the initially excited electrons, we can set up the following equations based on the energy levels:
- Let E_i be the energy of the initially excited state.
- Let E_f be the energy of the final state after the photon emission.
- The energy of the emitted photon can be expressed as: E_photon = E_i - E_f.
Given that the maximum photon energy is 13.4 eV, we can set up the equation:
E_i - E_f = 13.4 eV
Now, we need to find the possible values of n for the excited states. We can assume that the initial excited states correspond to n = 2, 3, and 4, as these are the first few excited states of hydrogen. Let's calculate the energies:
- For n = 2: E_2 = -13.6 eV / 2² = -3.4 eV
- For n = 3: E_3 = -13.6 eV / 3² = -1.51 eV
- For n = 4: E_4 = -13.6 eV / 4² = -0.85 eV
Now, we can find the energy differences:
- From n = 2 to n = 1: E_2 - E_1 = -3.4 eV - (-13.6 eV) = 10.2 eV
- From n = 3 to n = 1: E_3 - E_1 = -1.51 eV - (-13.6 eV) = 12.09 eV
- From n = 4 to n = 1: E_4 - E_1 = -0.85 eV - (-13.6 eV) = 12.75 eV
Thus, the principal quantum numbers of the initially excited electrons are likely n = 2, 3, and 4.
(b) Finding Maximum and Minimum Energies of Emitted Photons
Next, we need to determine the maximum and minimum energies of the initially emitted photons. The maximum energy of the emitted photon is already given as 13.4 eV. To find the minimum energy, we consider the lowest energy transition possible from the excited states.
For the minimum energy photon, we can look at the transition from n = 3 to n = 2:
E_photon(min) = E_3 - E_2 = (-1.51 eV) - (-3.4 eV) = 1.89 eV
Therefore, the maximum energy of the emitted photons is 13.4 eV, and the minimum energy is 1.89 eV.
Summary of Findings
- Principal Quantum Numbers: n = 2, 3, 4
- Maximum Energy of Emitted Photons: 13.4 eV
- Minimum Energy of Emitted Photons: 1.89 eV
This breakdown should clarify how to approach problems involving excited states and photon emissions in hydrogen. If you have any further questions or need more examples, feel free to ask!
