A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.77 nm. It then gives off a photon having a wavelength of 1100 nm. What is the final state of the hydrogen atom
I would do this.
Use E = hc/wavelenth. Plug in 93.77 nm and solve for E in joules, then use
dE = 2.180E-18*(1/1 - 1/x^2) and solve for x. I think you will get 6. That is where the electron has been lifted; i.e., to n = 6.
Then E = hc/wavelength and substitute 1100; solve for E. Substitute E into
dE = 2.180E-18*(1/x^2 - 1/36) and solve for x. I think that will be n = 3. That's the final state.
To determine the final state of the hydrogen atom, we need to consider the energy transitions that occur.
1. First, let's find the energy of the absorbed photon using the equation:
energy = Planck's constant × speed of light / wavelength
Plugging in the values:
energy = (6.626 × 10^-34 J·s) × (3.00 × 10^8 m/s) / (93.77 × 10^-9 m)
energy ≈ 2.091 × 10^-18 J
2. The absorbed photon causes the hydrogen atom to transition from the ground state (n=1) to an excited state (n > 1). We need to find the initial energy level of the hydrogen atom.
The energy of the initial state can be calculated using the equation:
energy_initial = -13.6 eV / n^2
Plugging in n=1:
energy_initial = -13.6 eV / (1)^2
energy_initial = -13.6 eV
3. To find the final state, subtract the energy of the absorbed photon from the initial energy level:
energy_final = energy_initial - energy
energy_final = -13.6 eV - (2.091 × 10^-18 J × (1 eV/1.602 × 10^-19 J))
energy_final ≈ -13.6 eV - 1.3 eV
energy_final ≈ -14.9 eV
4. Now we need to find the final principal quantum number (n_final) corresponding to the energy_final value. To do this, we use the same equation as before:
energy_final = -13.6 eV / n_final^2
Rearranging the equation and solving for n_final, we have:
n_final^2 = -13.6 eV / energy_final
n_final^2 ≈ (-13.6 eV) / (-14.9 eV)
n_final ≈ √0.912 ≈ 0.956
Since n must be a positive integer, we round up to the next closest integer value to get the final principal quantum number:
n_final = 1
Therefore, the final state of the hydrogen atom is the ground state, with the principal quantum number (n) being 1.
To determine the final state of the hydrogen atom, we need to understand the concept of photon absorption and emission in the context of atomic energy levels.
When an electron in a hydrogen atom absorbs a photon, it gains energy and moves to a higher energy level. Conversely, when an electron in a higher energy level loses energy, it emits a photon and moves to a lower energy level.
We can use the relationship between photon energy and wavelength given by the equation:
E = hc/λ
Where:
E = energy of the photon
h = Planck's constant (6.626 x 10^-34 J*s)
c = speed of light (3.00 x 10^8 m/s)
λ = wavelength of the photon
Given that the absorbed photon has a wavelength of 93.77 nm, we can calculate its energy:
E1 = (6.626 x 10^-34 J*s)(3.00 x 10^8 m/s) / (93.77 x 10^-9 m)
E1 ≈ 2.110 x 10^-18 J
Next, we need to find the energy of the emitted photon with a wavelength of 1100 nm:
E2 = (6.626 x 10^-34 J*s)(3.00 x 10^8 m/s) / (1100 x 10^-9 m)
E2 ≈ 1.804 x 10^-19 J
Since the emitted photon has less energy than the absorbed photon, it implies that the electron has transitioned from a higher energy level to a lower energy level.
Considering the energy levels of a hydrogen atom, we can conclude that the final state of the hydrogen atom is the n=2 energy level. This is known as the Balmer series, which corresponds to visible light.
In summary, the final state of the hydrogen atom after absorbing a photon with a wavelength of 93.77 nm and emitting a photon with a wavelength of 1100 nm is the n=2 energy level.