Final answer:
The electronic transitions in the hydrogen atom can be ranked from shortest to longest wavelength based on the energy difference, with the order being n = 2 to n = 1, n = 3 to n = 2, n = 16 to n = 7, and n = 6 to n = 1. The correct sequence is b, c, a, d.
Step-by-step explanation:
When ranking the electronic transitions in the hydrogen atom in order of increasing wavelength of the photon emitted, one should consider the energy difference between the initial and final energy levels. A greater energy difference means a shorter wavelength and vice versa. According to Bohr's model, the energy difference is given by:
E = -13.6 eV (1/n1² - 1/n2²)
Where E is the photon energy, n1 is the principal quantum number of the lower energy level, and n2 of the upper energy level, and 13.6 eV is the ionization energy of the hydrogen atom.
Using the relationship between energy (E), Planck's constant (h), and the speed of light (c) - E = hc/λ - we can find that a greater energy difference corresponds to a shorter wavelength (λ).
The transitions can be ordered from shortest to longest wavelength (or from largest to smallest energy difference):
- n = 2 to n = 1 (Largest energy difference, shortest wavelength)
- n = 3 to n = 2
- n = 16 to n = 7
- n = 6 to n = 1 (Smallest energy difference, longest wavelength)
Therefore, the correct order of increasing wavelength for the transitions given is b, c, a, d.