Final answer:
The correct wavelength of lines in the visible spectrum of hydrogen according to Johann Balmer's relation is given by the formula 364.6 n²/(n² - 4). There are four Balmer series lines visible, and calculations of transitions can be done using the Rydberg formula.
Step-by-step explanation:
The correct formula for the wavelength of lines in the visible spectrum of hydrogen according to Johann Balmer's relation is A) 364.6 n²/(n² - 4). This formula is derived from Balmer's equation which predicts the four wavelengths of the hydrogen Balmer series that are visible: 410.3 nm (n=6), 434.2 nm (n=5), 486.3 nm (n=4), and 656.5 nm (n=3). These wavelengths correspond to electronic transitions from higher energy levels to the second energy level (n=2).
The first line in the UV part of the spectrum is for the transition from n=3 to n=2, which is just outside the visible range. There are four Balmer series lines in the visible part of the spectrum. There are infinitely many lines in the UV, but as n increases, the lines get closer together and eventually converge to a limit.
For the question regarding a wavelength of 4.653 µm observed in a hydrogen spectrum for a transition that ends in the nf = 5 level, we can use the formula for the energy of the photon emitted during the transition, which is given by the difference in the energy levels. Since the final level (nf) is given, we need to calculate the initial level (ni) using the Rydberg formula for hydrogen's spectral lines.