An electron in Li2+ falls from n = 2 to the ground state. Calculate the wavenumber, (νˉ), of the emitted photon.
To calculate the wavenumber of the emitted photon, we need to use the Rydberg formula. The Rydberg formula relates the wavenumber (νˉ) to the initial and final energy levels of the electron. Here are the steps to calculate the wavenumber:
Step 1: Determine the initial and final energy levels of the electron.
In this case, the electron falls from n = 2 to the ground state. The ground state corresponds to n = 1, and the initial state corresponds to n = 2.
Step 2: Calculate the energy difference between the initial and final energy levels of the electron.
The energy difference can be calculated using the formula ΔE = E_initial - E_final. For an electron in a hydrogen-like atom, the energy levels are given by the formula: E = -13.6 eV / n^2, where n is the principal quantum number.
For the initial energy level (n = 2):
E_initial = -13.6 eV / (2^2) = -13.6 eV / 4 = -3.4 eV
For the final energy level (n = 1):
E_final = -13.6 eV / (1^2) = -13.6 eV
ΔE = -3.4 eV - (-13.6 eV) = 10.2 eV
Step 3: Convert the energy difference into joules.
1 eV = 1.602 × 10^-19 J
So, ΔE = 10.2 eV × (1.602 × 10^-19 J/eV) = 1.6364 × 10^-19 J
Step 4: Calculate the wavenumber (νˉ).
The wavenumber (νˉ) is given by the formula: νˉ = ΔE / h, where h is the Planck's constant (h = 6.626 × 10^-34 J·s).
νˉ = (1.6364 × 10^-19 J) / (6.626 × 10^-34 J·s) = 2.47 × 10^14 s^-1
Therefore, the wavenumber (νˉ) of the emitted photon is approximately 2.47 × 10^14 s^-1.