Final answer:
The question pertains to the physics concept of electron transitions within a hydrogen atom. It addresses the relationship between the de Broglie wavelengths of an electron, the wavelengths of emitted photons during transitions, and how these can be calculated for different energy states using Bohr's model and the Rydberg formula.
Step-by-step explanation:
The problem you're asking about concerns the transitions of an electron in the hydrogen atom and the de Broglie wavelengths associated with different energy states. When an electron transitions from an energized (excited) state to the ground state, it emits radiation. The wavelength of the emitted photon (denoted as ⋅n) is related to the energy difference between the nth state and the ground state. According to the Bohr model, the larger the orbit number (n), the less energy difference there is between adjacent energy states, meaning the emitted photon has a longer wavelength. In other words, ⋅n increases as n increases.
For large values of n, the de Broglie wavelength of an electron in the nth state (λn) and in the ground state (λg) can be related to the wavelength of the emitted photon during the transition. This relationship derives from Bohr's postulates and the de Broglie hypothesis which links the properties of particles to wavelengths. The de Broglie wavelength is inversely proportional to the momentum of an electron, and since the momentum is related to the energy states of an electron in an atom, we have a connection between the energy states, their respective momentum, and therefore their de Broglie wavelengths.
To consider the emitted wavelength (⋅n), we utilize the Rydberg formula which relates the wavelengths of photon emissions to the initial and final energy levels of the electron's transitions. The Lyman series is an example of this, where photons are emitted as the electron drops from higher energy levels back to the ground state (energy level 1). Using the given wavelengths in the Lyman series and energy of the ground state (-13.6 eV), one can calculate the energies of the excited states using the energy-wavelength relationship, E = hc/λ, where h is Planck's constant and c is the speed of light.