Final Answer:
The wavelength of light emitted during the transition from n=6 to n=2 is 4.55 × 10^-7 meters.
Step-by-step explanation:
To find the wavelength of light emitted during the transition from n=6 to n=2, we first calculate the energy difference using the formula ΔE = E_final - E_initial, where ΔE is the change in energy, E_final is the energy of the final state, and E_initial is the energy of the initial state.
The energy levels in hydrogenic atoms are given by the formula E = -(hcR) / n², where h is Planck's constant, c is the speed of light, R is the Rydberg constant, and n is the principal quantum number.
For n=6, the energy E_final = -(hcR) / 6², and for n=2, the energy E_initial = -(hcR) / 2². Subtracting E_initial from E_final gives the energy difference ΔE.
Using the energy-wavelength relationship for photons, where energy (E) is proportional to the inverse of wavelength (λ), we can employ the equation ΔE = hc / λ, where h is Planck's constant and c is the speed of light. Rearranging this equation allows us to solve for wavelength (λ), giving λ = hc / ΔE. Substituting the calculated ΔE value into this equation yields the wavelength of light emitted during the transition.
Substituting the calculated ΔE into the formula λ = hc / ΔE, where h is Planck's constant (6.626 × 10^-34 Js) and c is the speed of light (3.00 × 10^8 m/s), we find the wavelength to be 4.55 × 10^-7 meters, representing the light emitted during the transition from n=6 to n=2 in the hydrogenic atom.