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 10^{7} [(1 / n_{1 }^{2}) - (1 / n_{2} ^{2})]

For Balmer series, n_{1} = 2 and n_{2} = 3, 4, 5.....

The first line will be obtained for n_{2} = 3. So we will have

(1 / λ) = 1.09677 X 10^{7} [(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 n_{2} = ∞. So, we will have

(1 / λ) = 1.09677 X 10^{7} [(1 / 4) - ( 1 / ∞)]

= 1.09677 X 10^{7} [(1 / 4) - 0]

Solving for λ, we get

λ = 3.647 X 10^{-7} m

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