According to the quantum-mechanical model for the hydrogen atom, which of the following transitions would produce light with longer wavelength: 3p to 2s or 4p to 2s?
A) The 3p to 2s transition would produce light with longer wavelength.
B) The 4p to 2s transition would produce light with longer wavelength.
B) The 4p to 2s transition would produce light with longer wavelength. Remember, in quantum mechanics, "p" orbitals have higher energy than "s" orbitals. So, when an electron transitions from a higher energy orbital (4p) to a lower energy orbital (2s), the energy released corresponds to a longer wavelength of light. It's like going from high-pitched opera singing to low-pitched clown honking.
To determine which transition would produce light with a longer wavelength, we need to compare the energy levels of the initial and final states of each transition. In the quantum-mechanical model for the hydrogen atom, the energy levels are given by the formula:
E = -13.6 eV / n^2
where E is the energy, n is the principal quantum number, and -13.6 eV is the energy of the ground state (n=1).
For the 3p to 2s transition, the initial state (3p) has a principal quantum number of 3, and the final state (2s) has a principal quantum number of 2. Plugging these values into the formula, we get:
E_initial = -13.6 eV / (3^2) = -13.6 eV / 9
E_final = -13.6 eV / (2^2) = -13.6 eV / 4
For the 4p to 2s transition, the initial state (4p) has a principal quantum number of 4, and the final state (2s) has a principal quantum number of 2. Plugging these values into the formula, we get:
E_initial = -13.6 eV / (4^2) = -13.6 eV / 16
E_final = -13.6 eV / (2^2) = -13.6 eV / 4
Comparing the energy levels between the transitions, we see that the energy levels of the initial and final states for both transitions are the same. This means that the transitions will produce light with the same wavelength. Therefore, neither the 3p to 2s transition nor the 4p to 2s transition would produce light with a longer wavelength.
So, the answer is:
C) Neither transition would produce light with longer wavelength.
To determine which transition would produce light with a longer wavelength, we need to understand the energy levels involved.
In the quantum-mechanical model for the hydrogen atom, the energy levels are determined by the principal quantum number (n). The larger the value of n, the higher the energy level.
Additionally, each energy level has different sublevels, specified by the azimuthal quantum number (l). In this case, we have transitions involving 3p and 4p sublevels.
The rule to determine the energy difference between two energy levels is given by the formula ΔE = E_final - E_initial = Rh * (1/n_final^2 - 1/n_initial^2), where Rh is the Rydberg constant.
In this case, since we are comparing two different transitions, we can ignore the Rydberg constant and only focus on the ratios of the principal quantum numbers.
Comparing the transitions:
1. 3p to 2s transition:
- n_initial = 3
- n_final = 2
2. 4p to 2s transition:
- n_initial = 4
- n_final = 2
Now, let's calculate the energy difference for each transition:
1. ΔE(3p to 2s) = 1/2^2 - 1/3^2 = 1/4 - 1/9 = 5/36
2. ΔE(4p to 2s) = 1/2^2 - 1/4^2 = 1/4 - 1/16 = 3/16
The energy difference is inversely proportional to the wavelength of the emitted light. Therefore, the transition with the larger energy difference (and smaller absolute value) will produce light with a longer wavelength.
Comparing the energy differences, we can see that ΔE(3p to 2s) has a smaller absolute value (5/36), while ΔE(4p to 2s) has a larger absolute value (3/16).
Therefore, the 3p to 2s transition would produce light with a longer wavelength. Thus, the correct answer is A) The 3p to 2s transition would produce light with longer wavelength.