Answer
For the shortest wavelength, the electron falls to the n=1 state. Using the same formulas, the energy difference is E = 2.1175*10^-18 J and the wavelength is 93.8nm
Work Step by Step
The longest and shortest wavelengths of light emitted by electrons in a hydrogen atom transitioning from the n=6 state can be calculated using the formula for energy difference in atomic transitions and the relationship between energy and wavelength.
For the longest wavelength, the electron falls to the n=5 state. The energy difference is calculated as E = -R*(1/nf^2 – 1/ni^2), where R is the Rydberg constant (2.178*10^-18 J), nf is the final state (5), and ni is the initial state (6). This gives E = 2.662*10^-20J. The wavelength (WL) is then found using the formula WL = hc/E, where h is Planck's constant (6.626*10^-34) and c is the speed of light (3*10^8). This gives a longest wavelength of 7467.317806nm.
For the shortest wavelength, the electron falls to the n=1 state. Using the same formulas, the energy difference is E = 2.1175*10^-18 J and the wavelength is 93.8nm.
