The first member of the Balmer series of hydrogen atom has wavelength of 656.3 nm. Calculate the wavelength and frequency of the second member of the same series. Given, `c=3xx10^(8)m//s`.

Explanation

`(1)/(lambda)=R[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]`
Identifing transitions
Substitution, simplification and `lambda_(2)=486.1 nm`
Using `c=f lambda_(t)`
getting the value of `f=6.1715xx10^(14)xx10^(14)Hz`
Detailes answer :
`(1)/(lambda=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
for first member of Balmer series
`n_(1)=2 and n_(2)=3`
`rArr (1)/(656.3 nm)=R((1)/(2^(2))-(1)/(3^(2))) " "....(i)`
for second member of Balamer series
`m_(1) and 2 and n_(2) =4`
`rArr (1)/(lambda)=R((1)/(2^(2))-(1)/(4^(2)))`
dividing (i)(ii)
`(lambda)/(656.3 nm)=(((1)/(4)-(1)/(9)))/(((1)/(4)-(1)/(16)))`
`rArr lambda=656.3 xx (516xx4)/(36.12)`
=481.1 nm
also `C= vxx lambda` where v= frequyency
`rArr 3xx10^(8)=v xx486.1xx10^(-9)`
`rArr v=6.1715xx10^(14)Hz`