The radii of Bohr's orbit are directly proportional to

Explanation

Consider the electron revolving in the nth orbit around the hydrogen nucleus. Let m and -e be the mass and charge of the electron, r te radius of the orbit and v the linear speed of the electron.
According to Bohr's postulate, centripetal force acting on the electron = electrostatic force of attraction exerted on the electron by the nucleus.
`therefore (mv^(2))/(r) =(1)/(pi epsilon_(0))*(e^(2))/(r^(2)) " " ` ...(1)
where `epsilon_(0)` is the permittivity of free space.
`therefore v^(2) =(e^(2))/(4pi epsilon_(0) m r) " " ` ...(2)
According to Bohr's second postulate, the angular momentum of the electron,
`mvr=(nh)/(2pi) " " ` ...(3)
Where h is Planck's constant and n is the principal quantum number which takes integral values `1, 2, 3,..., ` etc.
`v=(nh)/(2pi m r)`
`therefore v^(2)=(n^(2)h^(2))/(4pi^(2)m^(2)r^(2)) " " ` ...(4)
Equating the right hand sides of Eqs. (2) and (4),
`(e^(2))/(4pi epsilon_(0)m r)=(n^(2)h^(2))/(4pi^(2)m^(2)r^(2)) " " therefore r=(epsilon_(0)n^(2)h^(2))/(pi me^(2))`
`therefore r=((epsilon_(0)h^(2))/(pi me^(2))) n^(2) " "` ...(5)
Since `epsilon _(0),h,m` and e are constants, it follows that `r prop n^(2), i.e., the radius of a Bohr orbit of the electron in a hydrogen atom is directly proportional to the square of the principal quantum number.