Final answer:
The final state of the hydrogen atom after absorbing a photon with a wavelength of 94.92 nm and emitting a 4.05e3 nm photon is likely a higher energy level beyond n=4, as the emitted wavelength doesn't correspond with the typical transitions between the lower energy levels in hydrogen.
Step-by-step explanation:
To determine the final state of a hydrogen atom after absorbing and then emitting photons, we can use the Bohr model of the hydrogen atom and its energy level transitions. When a ground state hydrogen atom absorbs a photon with a wavelength of 94.92 nm, it moves to an excited state. Since 94.92 nm corresponds to a transition in the Lyman series (which involves transitions to and from the ground state, n=1), this wavelength is close to the calculated wavelength for the transition from n=1 to n=4.
After absorption, the hydrogen atom emits a photon with a wavelength of 4.05e3 nm (4050 nm), which indicates a transition to a lower energy level. Given the energy states within the hydrogen atom, the emission of a 4.05e3 nm photon doesn't fit well with the transitions between the lower energy states (n=1 to n=5), which are generally in the visible to ultraviolet range. However, hydrogen's Brackett series or Pfund series can have transitions in the infrared, which can lead to such long wavelengths.
Given that the initial absorbed photon has a higher energy and corresponds to a transition to the n=4 state, and that the emitted photon has much lower energy (longer wavelength), it's likely that the final state is a higher energy level than n=4, but not a simple transition to the ground state as those have much shorter associated wavelengths. Therefore, it's plausible that the final state of the hydrogen atom is one of the higher energy levels beyond n=4. To provide a precise answer, though, one would need to calculate the exact energy levels using the Rydberg formula and compare them with the known spectral lines of hydrogen.