Answer:
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
