The wavelength of `H_(alpha)` line is Balmer series is 6563 Å. Compute the wavelength of `H_(beta)` line of Balmer series.

Explanation

Data: `lambda_(alpha) = 6563 Å`
`1/lambda = R(1/n_(1)^(2) -1/n_(2)^(2))`
For the `H_(alpha)` line of the Balmer series, `n_(1)=2` and `n_(2)=3`.
`therefore 1/lambda_(alpha) = R(1/2^(2) -1/3^(2)) =R(1/4-1/9)=(5R)/36`.........................(i)
For the `H_(lambda)` line of the Balmer series, `n_(1)=2` and `n_(2)=5`,
`1/lambda_(y) = R(1/2^(2) - 1/5^(2)) = R(1/4-1/25)= (21R)/100`...............(ii)
From eqs.., (i) and (ii), we have,
`lambda_(y)/lambda_(alpha) = (5R)/36 xx 100/(21R) = (5 xx 25)/(9 xx 21)`
`therefore lambda_(y) = 125/189 xx lambda_(alpha) = (125 xx 6563)/(189) = 4340 Å`
For the shortest wavelength `lambda_(m)` of the Lyman series, `n_(1)=1` and `n_(2) = infty, therefore 1/lambda_(m) = R(1/1^(2) - 1/infty^(2))=R`..................(iii)
From Eqs.(1) and (3), we have,
`lambda_(m)/lambda_(alpha) = (5R)/36 xx 1/R`
`therefore lambda_(m) = 5/36 lambda_(alpha) = 5/36 xx 6563 = 911.5 Å`