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 2630 nm. What is the final state of the hydrogen atom?
To determine the final state of the hydrogen atom, we need to understand the concept of energy levels and transition in atoms.
The hydrogen atom consists of a single electron orbiting a nucleus with one proton. The electron can exist in different energy levels or states. When an electron absorbs energy, it moves to a higher energy level, and when it emits energy, it moves to a lower energy level.
According to the given information, the hydrogen atom initially absorbs a photon with a wavelength of 93.77 nm. To find the corresponding energy of this photon, we can use the relationship between energy (E) and wavelength (λ) provided by 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 3.0 × 10^8 m/s).
- λ is the wavelength of the photon.
Plugging in the values, we can calculate the energy of the absorbed photon:
E = (6.626 × 10^-34 J*s × 3.0 × 10^8 m/s) / (93.77 × 10^-9 m)
E ≈ 2.106 × 10^-19 J
Now, the hydrogen atom emits a photon with a wavelength of 2630 nm. To determine the energy of this emitted photon, we use the same equation as before:
E = hc/λ
Plugging in the values:
E = (6.626 × 10^-34 J*s × 3.0 × 10^8 m/s) / (2630 × 10^-9 m)
E ≈ 7.558 × 10^-20 J
The difference between the initial and final energy levels of the hydrogen atom can be calculated by subtracting the energy of the emitted photon from the energy of the absorbed photon:
ΔE = E_final - E_initial
ΔE = 7.558 × 10^-20 J - 2.106 × 10^-19 J
ΔE ≈ -1.35 × 10^-19 J
The negative value for ΔE indicates that energy is being released or emitted by the hydrogen atom. This energy corresponds to the transition of the electron from a higher energy state to a lower energy state.
In hydrogen, transitions are described using spectral notation, which consists of four quantum numbers: n, l, m, and s. The principal quantum number (n) refers to the energy level of the electron, and the other three quantum numbers describe additional properties of the electron's orbit.
Since the hydrogen atom starts at the ground state (n = 1) and transitions to a lower energy state, we know the final principal quantum number (n) will be less than 1.
With the given information, we cannot determine the exact final state of the hydrogen atom without additional details. However, based on the energy release, we can infer that the hydrogen atom transitions to a lower energy level, likely n = 1 or n = 2.
To precisely determine the final state, we would need information about the remaining quantum numbers (l, m, and s) or further constraints on the system.