g The emission line with the longest wavelength. 2.) The absorption line with the shortest wavelength. 3.) The emission line with the highest energy. b 4.) The absorption line with the highest energy. a 5.) The emission line with the lowest frequency. a 6.) The line corresponding to the ionization energy of hydrogen.

The complete question is shown in the image attached to this answer.

a) 4

b) 2

c) 5

d) 2

e) 4

f) 3

Explanation:

It is a fact that shorter the wavelength implies greater energy. Among the emission lines, the greatest energy and shortest wavelength is 4.

From the same perspective as above, we can see that the greatest energy catapults the electron from energy level n=1 to n=4. This now corresponds to the shortest wavelength since energy is known to be inversely proportional to wavelength.  

Frequency is inversely related to wavelength when waves  are travelling at the same speed. Thus, a transition corresponding to the highest energy and shortest wavelength in the emission spectrum will have the highest frequency.

The emission line with the lowest energy is found at the line in which the electron moves from n=2 to n=1.

The ionization energy is the energy required to entirely remove the electron from the influence of the nucleus of the hydrogen atom. It corresponds to the transition n=1 to n=∞

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.