What is the shortest-wavelength line in the emission spectrum of the hydrogen atom?

The shortest wavelength line in the emission spectrum of the hydrogen atom is 364 nm.

Explanation:

The Balmer formula can be used to find out the wavelength which is as follow:

                      λ = b (n₂² / ( n₂² −4) )

where b = 364.56 nm and  n₂ = 3,4,5,6

It can be re-written as

ν˜ =  1 / λ

ν˜ = Rₐ ( 1/4 -1/n₂²)

Rₐ = 109,737 cm⁻¹ which is a Rydberg's constant.

For shortest wavelength, the energy would be maximum then:

ν˜ = 109737 (1/4 - 1/∞)

ν˜ = 109737 (1/4)

ν˜ =  27434 cm⁻¹

As ν˜ =  1 / λ and also λ = 1/ ν˜

λ = 1/ 27434

λ =  364 nm

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.