Show that the radius of the orbit in hydrogen atom varies as `n^(2)`, where it is the principal quantum number of the atom.

Explanation

We know that when an electron revolves in a stable orbit, the centripetal force is provided by electrostatic force of attraction acting on it due to a proton present in the nucleus.
`therefore " " (m v_(n)^(2))/(r_(n)) = (1)/(4pi in_(0)) .(e^(2))/(r_(n)^(2)) or v_(n) = (nh)/( 2pi m r_(n))" ".....(i)`
and from Bohr.s quantum condition , we have .
`m v_(n) r_(n) =(n h)/(2 pi) or v_(n) =(n h)/( n h)/(4 pi in_(0) m r_(n))" "........(ii) `
Squaring (ii) and then equating it with (i) , we get .
`(n^(2)h^(2))/(4 pi^(2) m^(2) r_(n)^(2)) = (e^(2))/(4 pi in_(0) m.r_(n)) rArr r_(n) = (n^(2) h^(2))/(4pi^(2) m^(2)) xx (4 pi in_(0) m)/(e^(2)) = ( in_(0)h^(2))/( pi m e^(2)) = n^(2)`
In stable orbit of hydrogen atom n = 1 and then radius of 1st orbit is called Bohr.s radius `a_(0)` . Obviously ` a_(0) = (in_(0) h^(2))/(pi m e^(2))`