Find the wavelength of the spectral line corresponding to the transition in hydrogen from n = 6 state to n = 3 states. (Ans: 1094 nm)



Data Given:


Excited state = n = 8

Lower state = p = 1


∴ Rydberg constant = R = 1.097 ｘ 10⁷ m⁻¹


To Find:

The wavelength of the spectral line = Î» =  ?



>

Solution:

Using the Rydberg formula for electron transition from excited state p to lover state n

`\frac {1}{λ}` = R`(\ frac {1}{p²} - frac {1}{n²})`

by putting values

`\frac {1}{λ}` = 1.097 ｘ 10⁷ m⁻¹ `(\ frac {1}{3²} - frac {1}{6²})`

`\frac {1}{λ}` = 1.097 ｘ 10⁷ m⁻¹ `(\ frac {1}{9} - frac {1}{36})`

`\frac {1}{λ}` = 1.097 ｘ 10⁷ m⁻¹ `\frac {4 -1}{36}`

`\frac {1}{λ}` = 1.097 ｘ 10⁷ m⁻¹ `\frac {3}{36}`

`\frac {1}{λ}` = `\frac {1.097 ｘ 10⁷ m⁻¹}{12}`


by flipping the fraction on both sides of the equation we get


λ = `\ frac {12}{1.097 ｘ 10⁷ m⁻¹}`


or


λ = 10.938 ｘ 10⁻⁷ m


or


λ = 1093.8ｘ 10⁻⁹ m


λ = 1094 nm ------------Ans.



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