The time of revolution of an electron around a nucleus of charge `Ze` in `n`th Bohr orbit is directly proportional to

Options

A. `n`
B. `(n^(3))/(Z^(2))`
C. `(n^(2))/(Z)`
D. `(Z)/(n)`

Answer

B

Explanation

`T = (2pi r)/(v) , r =` radius of `nth` orbit `= (n^(2)h^(2))/(pi mZe^(2))`
`v =` speed of `bar(e)` in nth orbit `= (ze^(2))/(2 epsilon_(0)nh)`
`:. T = (4epsilon_(0)^(2)n^(3)h^(3))/(mZ^(2)e^(4)) rArr T prop (n^(2))/(Z^(2))`