Calculate the wavelength of first and last line in the Balmer series of Hydrogen spectrum.

Here we will use the following relation 

(1 / λ) = 1.09677 X 10⁷ [(1 / n₁ ²) - (1 / n₂ ²)]

For Balmer series, n₁ = 2 and n₂ = 3, 4, 5.....

The first line will be obtained for n₂ = 3. So we will have

(1 / λ) = 1.09677 X 10⁷ [(1 / 4) - ( 1 / 9)]

Solving for λ, we get 

λ = 6.563 X 10^-7 m

Similarly, last line of Balmer series will be obtained for n₂ = ∞. So, we will have

(1 / λ) = 1.09677 X 10⁷ [(1 / 4) - ( 1 / ∞)]

= 1.09677 X 10⁷ [(1 / 4) - 0]

Solving for λ, we get 

λ = 3.647 X 10^-7 m