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.

The angular momentum of an electron in a particular orbit of Li^(2+) i

1.	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.
Answer» Solution :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`