The angular momentum of an electron in a Bohr's orbit of `He^(+)` is `3.1652xx10^(-34)` kg-`m^(2)`/sec. What is the wave number in terms of Rydberg constant (R ) of the sepectral line emitted when an electron falls from this level to the first excited state.l [ Use h`=6.626xx10^9-34)` Js]

Answer

`R((8)/(9))`

Explanation

`mvr=(nh)/(2pi)=3.16xx10^(-34)`,`n=(2pi)/(h)xx3.16xx10^(-34)~~3`
`barV=(1)/(lambda)=R[(1)/(n_(1))-(1)/(n_(2)^(2)]]=R[(1)/(1^(1))-(1)/(3^(2)]] =(8R)/(9)`