A tube containing hydrogen atoms in the n=3 shell does not absorb visible light. The longest wavelength absorbed is 1875 nm. What state is the hydrogen atom in after absorbing a 1875 nm photon?
1/wavelength = R(1/n^2initial - 1/n^2final)
R is the Rydberg constant.
n final is the upper state
n initial is the lower state, in this case 3.
Plug in the numbers and solve for n final. Post your work if you get stuck.
To determine the state of the hydrogen atom after absorbing a 1875 nm photon, we need to understand the energy levels and transitions in the hydrogen atom.
In the hydrogen atom, the energy levels are labeled using the principal quantum number (n). The n=1 level is the lowest energy level, followed by n=2, n=3, and so on.
When a hydrogen atom absorbs a photon, it undergoes a transition from its initial energy level to a higher energy level. The energy of a photon can be calculated using the equation:
E = (hc) / λ
Where:
E is the energy of the photon,
h is Planck's constant (approximately 6.626 × 10^(-34) J·s),
c is the speed of light (approximately 3.00 × 10^8 m/s),
λ is the wavelength of the photon.
Given that the longest wavelength absorbed is 1875 nm, we can calculate the energy of the absorbed photon using the equation above.
First, let's convert the wavelength to meters:
λ = 1875 nm = 1875 × 10^(-9) m
Now we can calculate the energy of the photon:
E = (6.626 × 10^(-34) J·s * 3.00 × 10^8 m/s) / (1875 × 10^(-9) m)
E ≈ 3.355 × 10^(-19) J
Next, we need to compare this energy to the energy difference between different energy levels in the hydrogen atom. The energy difference between any two hydrogen energy levels can be calculated using the equation:
ΔE = -13.6 eV * (1/n_final^2 - 1/n_initial^2)
Where:
ΔE is the energy difference,
-13.6 eV is the ionization energy of hydrogen.
We can plug in the values for the energy levels to determine the energy difference:
For the n=3 initial energy level:
ΔE = -13.6 eV * (1/n_final^2 - 1/3^2)
To find the final energy level (n_final) that corresponds to the absorbed photon's energy, we equate ΔE to the absorbed photon's energy:
3.355 × 10^(-19) J = -13.6 eV * (1/n_final^2 - 1/3^2)
Now, we need to solve for n_final. Rearranging the equation and substituting the values:
1/n_final^2 = 1/9 - (3.355 × 10^(-19) J / -13.6 eV)
1/n_final^2 ≈ 0.111111
n_final^2 ≈ 9
n_final ≈ 3
Since the n_final value is approximately 3, it means that the absorbed photon corresponds to a transition within the n=3 shell. Therefore, the hydrogen atom remains in the n=3 shell after absorbing the 1875 nm photon.