1.

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) = 109677 cm^(-1), c = 3 xx 10^(8) m s^(-1), Z = 3)`

Answer» Correct Answer - A::B::C::D
`(1)/(lambda) = bar v = R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
`= 1.09678 xx 10^(7)m^(-1)[(1)/(1^(2)) -(1)/(2^(2))]`
`= 1.09678 xx 10^(7)m^(-1)(1 - (1)/(4))`
`lambda = (4)/(1.09678 xx 10^(7) xx 3)m^(-1) = 1.2157 xx 10^(-7)m`
`
c/lambda=(3xx10^(8)ms^(-1))/(1.2157xx10^(-7)m)=2.4677xx10^(15)s^(-1)`
`E = hv = 6.6625 xx 10^(-14) Js xx 2.4677 xx 10^(15) s^(-1) = 1.63485 xx 10^(18) J` [ `E = hv = 6.6625 xx 10^(-14) Js xx 2.4677 xx 10^(15) s^(-1) = 1.63485 xx 10^(18) J` [Energy for comes- ponding line spetral of `Li^(2+)]` = [Energy for hydrogen atom `xx Z^(2)]`
`1.63485 xx 10^(-10) J xx 3^(2)`
` = 1.471.365 xx 10^(-17) J`


Discussion

No Comment Found

Related InterviewSolutions