Consider `3rd` orbit of `He^(+)` (Helium) using nonrelativistic approach the speed of electron in this orbit will be (given `K = 9 xx 10^(9)` constant `Z = 2` and `h` (Planck's constant) `= 6.6 xx 10^(-34)Js`.)

Options

A. `2.29xx10^(6)ms^(-1)`
B. `1.46xx10^(6)ms^(-1)`
C. `0.73xx10^(6)ms^(-1)`
D. `3xx10^(8)ms^(-1)`

Answer

B

Explanation

Energy of electron in 3rd orbit of `H^(+)` is
`E_(3)=-13.6xx(Z^(2))/(n^(2))eV=-13.6xx(4)/(3^(2))eV`
`=-13.6xx(4)/(9)xx1.6xx10^(-19)J`
From Bohr's model, `E_(3)=-KE_(3)=-(1)/(2)my^(2)`
`rArr(1)/(2)9.1xx10^(-31)xxv^(2)=-13.6xx(4)/(9)xx1.6xx10^(-19)`
`rArrv^(2)=(13.6xx4xx16xx10^(-19)xx2)/(9.1xx10^(-31)xx9)`
`rArrv^(2)=2.12xx10^(12)rArrv=1.46xx10^(6)ms^(-1)`