The wavelength of the first line in the hbalmer series is `656 nm ` .Calculate the wavelength of the second line and the limeting line in the Ralmerseries

Explanation

Wavelength of the second line
According to Rydbery formula
`(1)/(lambda) = R_(H)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
For the first line `n_(1) = 2,n_(2)= 3 and lambda = 656 nm`
`(1)/(656) = R_(H) ((1)/(2^(2)) - (1)/(3^(2)))`
` R_(H) = ((1)/(4) - (1)/(9)) = R_(H) xx (5)/(36) = (5R_(H))/(36)`....(i)
For the second line `n_(1) = 2,n_(2) = 4`
`(1)/(lambda) = R_(H) ((1)/(2^(2)) - (1)/(4^(2))) = R_(H)((1)/(4) - (1)/(16))`
`=R_(H) xx(3)/(16) = (3R_(H))/(16)`.....(ii)
Dividing equation (i) by equation (ii),we get
`(lambda)/(656) = (5)/(36) xx (16)/(3)`M
`or lambda = (656 xx 5xx 16)/(36 xx 3) = (52450)/(108) = 485.9 m`
IN balmer series the limiting line coreesponding to transition from `n = oo "to" n = 2` therefore
`n_(1) = 2 and n_(2) = oo`
`therefore (1)/(lambda) = R_(H)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
`(1)/(lambda) = 109677 [(1)/(2^(2)) - (1)/(oo^(2))] = 109677 xx (1)/(4) cm^(-1) = 27419.25 cm^(-1)`
Wavelength lambda = `(1)/(27419.25) = 3.47 xx 10^(-5) = 3647 Å`