Calculate frequency energy and wavelength of the radiation corresponding to the speciral line of the lowest frequency in lyman series in the spectrum of a hydrogen atom .Also calculate the energy for the coresponding line in the spectrum of `Li^(2+).(R_(H) = 109.677 cm^(-1), c = 3 xx 10^(6) m s^(-1), Z = 3)`

Explanation

In Lyman series `n_(1) = 1 and n_(2) = 2` (coreresponding lowest frequency region )
`(1)/(lambda) = bar v = R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] = 1.09677 xx 10^(7) m^(-1)[(1)/(1^(2))- (1)/(2^(2))]`
`= 1.09677 xx 10^(7) m^(-1)(1 - (1)/(4)) = (1.9677 xx 10^ (7) xx 3)/(4) cm `
`:. lambda = (4)/(1.09677 xx 10^(7) xx 3 ) = 1.215 xx 10^(7) m`
`"Frequency" v = (c )/(lambda) = (3 xx 10^(8) ms^(-1))/(1.215 xx 10^(-7) m) = 2.4677 xx 10^(15)s^(-1)`
Energy of radiation `E = hv`
`:. E = 6.625 xx 10^(-34) Js xx 2.4677 xx 10^(15)s^(-1)`
`= 1.6348 xx 10^(-16) J`
energy for corresponding line in the spectral of `Li^(2+)` = Energy for hydrogen atom `xx Z^(2) = 1.6348 xx 10^(-18)J xx 3^(2) = 1.471 xx 10^(-17)J`