An electron in the n = 10 state of a hydrogen atom emits a photon with energy of 6.54×10–20 J. What is the final state of the electron
To determine the final state of the electron, we need to understand the energy levels of the hydrogen atom and how photons are emitted.
In hydrogen atoms, the energy levels are described by the principle quantum number (n). The energy of an electron in a hydrogen atom is given by the formula:
E = -13.6 eV / n^2
where E is the energy of the electron in electron volts (eV) and n is the principal quantum number.
To find the final state of the electron, we need to determine the final principal quantum number (n) by using the energy of the emitted photon.
The energy of a photon (E_photon) is related to the energy difference between two energy levels in the atom. The relationship is given by:
E_photon = E_initial - E_final
where E_initial is the energy of the initial state and E_final is the energy of the final state.
First, convert the energy of the photon from joules to electron volts (eV). 1 electron volt is equal to 1.60 × 10^(-19) Joules.
E_photon = 6.54 × 10^(-20) J * (1 eV / 1.60 × 10^(-19) J) = 0.409 eV
Next, substitute the known values into the equation for the photon energy:
0.409 eV = (-13.6 eV / 10^2) - (-13.6 eV / n^2)
Simplify the equation:
0.409 eV = -1.36 eV - (-13.6 eV / n^2)
Rearrange the equation to solve for n:
-13.6 eV / n^2 = 0.409 eV + 1.36 eV
-13.6 eV / n^2 = 1.769 eV
Now, solve for n by isolating it:
n^2 = -13.6 eV / 1.769 eV
n^2 ≈ 7.686
Taking the square root of both sides:
n ≈ √7.686
n ≈ 2.77
Since n is a principal quantum number that must be a positive integer, the final state of the electron (n_final) would be rounded to the nearest whole number, which is 3.
Therefore, the final state of the electron is n = 3.
To find the final state of the electron, we need to determine the energy difference between the initial state (n = 10) and the final state.
The energy of a photon emitted or absorbed by an electron transitioning between energy levels in a hydrogen atom can be calculated using the equation:
E = -13.6 eV * (1/n_initial^2 - 1/n_final^2)
Where E is the energy of the photon, n_initial is the initial state, and n_final is the final state.
Let's rearrange the equation to solve for n_final:
E = -13.6 eV * (1/n_initial^2 - 1/n_final^2)
Rearranging the equation, we get:
1/n_final^2 = 1/n_initial^2 - E / (-13.6 eV)
Now, let's substitute the values given:
n_initial = 10
E = 6.54×10^(-20) J
Since the energy unit given is in joules, we need to convert it to electron volts (eV). The conversion factor is:
1 eV = 1.602 × 10^(-19) J
Therefore:
E = 6.54×10^(-20) J * (1 eV / 1.602 × 10^(-19) J) ≈ 0.408 eV
Now, let's plug in the values:
1/n_final^2 = 1/10^2 - 0.408 eV / (-13.6 eV)
Simplifying:
1/n_final^2 = 0.01 - 0.408/(-13.6)
1/n_final^2 = 0.01 + 0.030
1/n_final^2 = 0.040
Taking the square root of both sides:
1/n_final ≈ ± 0.2
To determine the final state, we need to find a whole number for n_final. Taking the reciprocals of the above values, we get:
n_final ≈ ± 5
Since the energy difference is associated with the emission of a photon, the final state is expected to have a lower energy than the initial state. Therefore, we can conclude that the final state of the electron is n = 5.