To determine which excited state of hydrogen corresponds to the photons that can pass through a filter allowing only wavelengths greater than 800 nm, we need to analyze the energy levels of the hydrogen atom and the relationship between energy and wavelength. Let's break this down step by step.
Understanding Energy Levels in Hydrogen
The energy levels of a hydrogen atom can be calculated using the formula:
E_n = -13.6 eV / n²
Here, E_n is the energy of the nth level, and n is the principal quantum number (1 for the ground state, 2 for the first excited state, and so on).
Calculating Energy Levels
Let's calculate the energy for the first few excited states:
- For the ground state (n=1): E_1 = -13.6 eV
- For the first excited state (n=2): E_2 = -13.6 eV / 2² = -3.4 eV
- For the second excited state (n=3): E_3 = -13.6 eV / 3² ≈ -1.51 eV
- For the third excited state (n=4): E_4 = -13.6 eV / 4² = -0.85 eV
- For the fourth excited state (n=5): E_5 = -13.6 eV / 5² = -0.544 eV
Photon Emission and Wavelength
When an electron transitions from a higher energy level to a lower one, it emits a photon. The energy of the emitted photon can be calculated using:
E_photon = E_initial - E_final
The wavelength of the emitted photon can be found using the equation:
λ = hc / E_photon
Where h is Planck's constant (4.1357 x 10^-15 eV·s) and c is the speed of light (3 x 10^8 m/s). For our calculations, we can use the given value of hc = 1240 eV·nm.
Finding the Wavelengths for Each Transition
Now, let's consider the transitions from the excited states to the ground state:
- From n=2 to n=1:
- E_photon = -3.4 eV - (-13.6 eV) = 10.2 eV
- λ = 1240 eV·nm / 10.2 eV ≈ 121.57 nm
- From n=3 to n=1:
- E_photon = -1.51 eV - (-13.6 eV) = 12.09 eV
- λ = 1240 eV·nm / 12.09 eV ≈ 102.4 nm
- From n=4 to n=1:
- E_photon = -0.85 eV - (-13.6 eV) = 12.75 eV
- λ = 1240 eV·nm / 12.75 eV ≈ 97.25 nm
- From n=5 to n=1:
- E_photon = -0.544 eV - (-13.6 eV) = 13.056 eV
- λ = 1240 eV·nm / 13.056 eV ≈ 95.03 nm
Considering Higher Excited States
Next, let's consider transitions from n=3, n=4, and n=5 to n=2:
- From n=3 to n=2:
- E_photon = -1.51 eV - (-3.4 eV) = 1.89 eV
- λ = 1240 eV·nm / 1.89 eV ≈ 655.06 nm
- From n=4 to n=2:
- E_photon = -0.85 eV - (-3.4 eV) = 2.55 eV
- λ = 1240 eV·nm / 2.55 eV ≈ 486.27 nm
- From n=5 to n=2:
- E_photon = -0.544 eV - (-3.4 eV) = 2.856 eV
- λ = 1240 eV·nm / 2.856 eV ≈ 434.24 nm
Identifying the Correct Excited State
Now, we need to find the excited state that can emit photons with a wavelength greater than 800 nm. The only transition that results in a wavelength greater than 800 nm is from n=5 to n=4, which we haven't calculated yet:
- From n=5 to n=4:
- E_photon = -0.544 eV - (-0.85 eV) = 0.306 eV
- λ = 1240 eV·nm / 0.306 eV ≈ 4052.61 nm (which is not greater than 800 nm)
After reviewing all the calculations, we find that the only transitions that yield wavelengths greater than 800 nm are from the 5th excited state. Therefore, the answer is:
Final Answer
5th excited state
