Calculate the velocity of electron in Bohr's first orbit of hydrogen atom. How many times does the electron go in Bohr's first orbit in one second ?

Answer

`2.18xx10^(6) m s^(-1)`

Explanation

According to Bohr's quantisation of angular momentum, `mvr=(nh)/(2pi) orv=(nh)/(2 pi mr)`
For the first orbit `n=1, m=9.1xx10^(-31) "kg", " "r=0.53 Ã… or 0.53xx10^(-10)m`
`h=6.626xx10^(-34) "kg" m^(2)s^(-1)`
`:. " "v=(1xx(6.626xx10^(-34) "kg"m^(2)s^(-1)))/(2xx3.143xx(9.1xx10^(-31) "kg")xx(0.53xx10^(-10)m))=2.18xx10^(6) m s^(-1)`.