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. `1.46xx10^(6)m//s`
B. `0.73xx10^(6)m//s`
C. `3xx10^(8)m//s`
D. `2.92xx10^(6)m//s`

Answer

A

Explanation

Here, Z=2, n=3
for H-like atoms, `v=Z/nxx2.2xx10^(6)m//s`
`v=2/3xx2.2xx10^(6)=(4.4xx10^(6))/3=1.46xx10^(6)m//s`