The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of `3`:

Options

A. `10^(20)`
B. `10^(19)`
C. `10^(17)`
D. `10^(15)`

Answer

D

Explanation

`mvr = (h)/(2pi)` (for first orbit)
`rArr m omega r^(2) = (h)/(2pi) rArr m xx 2pi v xx r^(2) = (h)/(2pi)`
`rArr v = (h)/(4pi^(2)mr^(2))`
`= (6.6 xx 10^(-34))/(4(3.14)^(2) xx 9.1 xx 10^(-31) xx (0.53 xx 10^(-10))^(2))`
`6.5 xx 10^(15) (rev)/(sec)`