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

Explanation

Given `lamda _(alpha ) = 6563 A^(@) = 6.536 xx10 ^(-7) m`
` lamda _( beta) = ? , f_(B) = ? , C=3xx10 ^(8) ms^(-1)`
wkt `(1)/( lamda ) = R ((1)/(n_(1)^(2))-(1)/(n_(2)^(2))).` for I member of balmer series ,` n _(1) =2 , n _(2) =3`
` therefore (1)/(6xx653xx10 ^(-7))=R((1)/(2^2)-(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)/(lamda _(2)) = 1.097 xx10 ^(7) ((1)/(2^2)-(1)/(4^2))`
`i.e., (1)/(lamda _2)=(1.097 xx10 ^(7)xx12 )/(64 )`
hence `lamda _(2) =(64) /(1.097xx12xx10 ^(7))m`
` ie., lamda _(2) = 4.8617xx10 ^(7) m`
or ` lamda _(2) =4561.7 A^(@)`
and frequency `gamma _(2) =(C )/(lamda_2)=(3xx10 ^(8))/( 4.8617 x10 ^(-7))`
`ie., gamma _(2) = 0.6171 xx10 ^(15) Hz = 6.171 xx 10 ^(14) Hz.`