To tackle the problem of the electron transition in a hydrogen atom, we need to understand a few key concepts about energy levels, photon emission, and the relationship between energy and wavelength. The transition you're describing involves an electron moving from a higher energy state to a lower one, which results in the emission of a photon. Let's break this down step by step.
Understanding Energy Levels in Hydrogen
In a hydrogen atom, the energy levels are quantized and can be described using the formula:
E_n = -13.6 eV / n²
where E_n is the energy of the level and n is the principal quantum number. The ground state corresponds to n = 1, and the energy of this state is -13.6 eV.
Calculating the Energy of the Excited State
Given that the excitation energy with respect to the ground state is 10.2 eV, we can find the energy of the excited state:
E_2 = E_1 + 10.2 eV
Since the ground state energy (E_1) is -13.6 eV, we can calculate:
E_2 = -13.6 eV + 10.2 eV = -3.4 eV
Finding the Quantum Number of the Excited State
Now, we can determine the quantum number n for this energy level:
-3.4 eV = -13.6 eV / n²
Rearranging gives:
n² = 13.6 eV / 3.4 eV = 4
Thus, n = 2. This means the electron transitions from n = 2 to n = 1.
Calculating the Wavelength of the Emitted Photon
Next, we can find the wavelength of the emitted photon using the energy of the photon emitted during the transition. The energy of the photon can be calculated using the formula:
E_photon = E_initial - E_final
Here, E_initial is the energy of the excited state and E_final is the energy of the ground state:
E_photon = (-3.4 eV) - (-13.6 eV) = 10.2 eV
Relating Energy to Wavelength
The energy of a photon is also related to its wavelength by the equation:
E_photon = hc / λ
Where:
- h is Planck's constant (6.626 x 10^-34 J·s)
- c is the speed of light (3.00 x 10^8 m/s)
- λ is the wavelength in meters
First, we need to convert the energy from electronvolts to joules:
10.2 eV = 10.2 x 1.6 x 10^-19 J = 1.632 x 10^-18 J
Calculating the Wavelength
Now we can rearrange the energy-wavelength equation to solve for wavelength:
λ = hc / E_photon
Substituting the values:
λ = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s) / (1.632 x 10^-18 J)
Calculating this gives:
λ ≈ 1.22 x 10^-6 m or 1220 nm
Final Thoughts
In summary, the transition of the electron from the second energy level to the ground state in a hydrogen atom results in the emission of a photon with a wavelength of approximately 1220 nm. This wavelength falls in the infrared region of the electromagnetic spectrum. Understanding these transitions and the associated energies is fundamental in quantum mechanics and atomic physics.
