How many times dies the electon go round the first Bohr`s orbit of hydrogen in one second ?

Explanation

For H-atom : `u_1 = ( 2pie^2)/h , r_1 = h^2 /(4 pi^2 me^2)`
` therefore` Circumference fo I Bohr orbit ` = 2 pi r_1`
` = ( 2 pi h^2 )/( 4 pi me^2) = h^(2)/(2pime^(2))` ` therefore` No . of round by electron in I orbit
` = u_1 /(2 pir_1) =( 2 pe^2 xx 2 pi me^2) /(hxx h^2) = ( 4 pi^2me^4)/(h^3)`
` = 4 xx (3 . 14 )^2 xx 9.108 xx 10^(-28) xx ((4. 803 xx 10^(-10))^4)/((6.626 xx 10^(-27)))^3`
`= 6.57 xx 10^(15)`.