What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is `-3.4e V`?

Explanation

Energy of electron in `n^(th)` Bohr orbit of hydrogen atom is given by,
`E= - ( 13.6)/( n^(2))eV`
Hence,` - 3.4 = - ( 13.6)/( n^(2))`
or `n^(2) = 4`
or `n =2 `
The angular momentum of an electron in `n^(th)` orbit is given a s `L = ( nh )/( 2pi )` . Putting `n= 2 `, we obtain
`L =( 2h )/( 2pi ) = ( h )/( pi ) rArr T = ( n^(3) h^(3))/( 4pi^(2) k^(2) Z^(2) e^(4) m )`
Thus,` T prop n^(3)` or ` ( T_(1))/( T_(2)) = ( n_(1)^(3))/( n_(2)^(3))`
As `T_(1) 8 T_(2)`, the above relation gives
`((n_(1))/(n_(2)))^(3) = 8` or `n_(1) = 2n_(2)`
Thus the possible values of `n_(1)` and `n_(2)` are
`n_(1) 2, n_(2) =1, n_(1) = 4, n_(2) = 2, n_(1) = 6, n _(2) = 3 ` and so on `"........."`.