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

Given `lambda_(alpha)=6563 Å = 6.536 xx 10^(-7) m`
`lambda_(beta)= ? , f_(beta)=?`
`c=3xx10^(8) ms^(-1)`
wkt `(1)/(lambda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` for I member of balmer series `n_(1)=2, n_(2)=3`
`therefore (1)/(6xx563xx10^(-7))=R((1)/(n_(1)^(2))-(1)/(3^(2)))`
ie `(10^(7))/(6.563)=R(5)/(36)`
Hence R= `36/5 xx(10^(7))/(6.563)`
i.e R= `1.097 xx10^(7) m^(-1)`
For II member of balmer series
`x_(1) =2, x_(2) =4`
`therefore (1)/(lambda_(2))=1.097 xx10^(7)(1)/(2^(2))-(1)/(4)^(2)`
Hence `lambda_(2)=(64)/(1.097 xx12xx10^(7))`m
ie `lambda_(2)=4.8617 xx10^(-7)` m
and frequency `gamma_(2)=(C )/(lambda_(2))=(3xx10^(8))/(4.8617xx10^(-7))`
ie `gamma_(2)=0.6171 xx10^(15) Hz=6.171 xx10^(14)` Hz