The angular momentum of an electron in a particular orbit of `Li^(2+)` ion is 5.2728 `xx 10^(-34)` kg `m^(2) //` sec. Calculate the frequency of the spectral line when electron falls from this level to the level where angular momentum of electron will be `3.1636xx 10^(-34) kg`-`m^(2)//` sec.

Explanation

Angular momentum of the electron `= 5.2728 xx 10^(-34) kgm^(2) //` sec
`( n_(1) h )/( 2pi ) = 5.272 xx 10^(-34)`
`n_(1) = 4.99 ~~ 5 `
Angular momemtum of electron after transition ` =3.1636 xx 10^(-34) kgm^(2) //` sec
`(n_(1) h )/( 2pi ) = 3.1636 xx 10^(-34)`
`n_(2) = 2.999 ~~ 3 `
`:.` the electron makes transition from `. 5 rarr 3 `
`v = c bar( v ) = c xx R_(h ) xx Z^(2) [ ( 1)/( n_(1)^(2) ) - ( 1)/( n _(2)^(2)) ]`
`v= 3 xx 10^(10) xx 109678xx 3^(2) xx [ ( 1)/( 3^(2)) - ( 1)/( 5^(2))] = 2.1058 xx 10^(15) Hz`