Consider `3^(rd)` orbit of `He^(+)` (Helium), using non-relativistic approach, the speed of electron in this orbit will be (given `K=9xx10^(9)` constant, Z=2 and h (Planck's constant) `=6.6xx10^(-34)J-s)`

Options

A. `1.46 xx10^6 m//s`
B. `0.73 xx10^6 m//s`
C. `3.0 xx10^8`
D. `2.92 xx10^6 m//s`

Answer

A

Explanation

Energy is `n^(th)` orbit –
`E_n=(-13.6 Z^2)/(n^2)eV implies E_3 = (-13.6 xx(2)^2)/((3)^2)eV`
`(KE)_3 =-E_3=(13.6 xx4)/9 eV`
`=1/2 mv_3^2 = (13.6 xx4)/9 eV`
`impliesv_3=sqrt((13.6xx4xx1.6 xx10^(-19)xx2)/(9xx9.1 xx10^(-31)))=146xx10^6 m//s`