If the value of Rydberg constant of hydrogen is `109737 cm^(-1)` , determine the longest and the shortest wavelength of the Balmer series.

Explanation

If the wavelength of Balmer series be `lambda` , then
`1/(lambda)=R(1/(2^2)-1/(n^2)) " " [n=3,4,5...oo, R="Rydberg constant"]`
substituting the minimum value of n,i.e., n=3,we get,
`1/lambda=R(1/(2^2)-1/(3^2))=Rxx5/(36)`
`or, lambda=(36)/(5R)=(36)/(5xx109737)"Å"=6561"Å"`
This is the longest wavelength.
Again , substituting the maximum value of `n,i.e., n=oo,` we get
`1/(lamda)=R(1/2^2-1/oo)=Rxx1/4`
or, `lambda=4/R=(4)/(109737)cm=(4xx10^8)/(109737)"Å"=3645"Å"`
This is the shorest wavelength.