According to the bohr model of the atom, which electron transition would correspond to the shortest wavelength line in the visible emission spectra for hydrogen?

Explanation:

According to Bohr's model, electrons are travelled in defined circular orbit around the nucleus. The orbits of different energy level is labelled by quantum numbers (n). Electrons can jump different energy levels, from higher to lower energy level by releasing energy.

In the visible emission spectrum of hydrogen, there are mainly five series which we are studied in regular. Series are: Lyman series(n=1), Balmer series(n=2), Paschen series(n=3), Bracket series(n=4), Pfund series(n=5).

Wavelength is inversely proportional to the quantum number(n). So, we can conclude that greater the value of n shortest the wavelength. According to above spectral series Pfund series(n=5) have shortest wavelength.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.