Making use of a formula of the foregoing problem, estimate the time during which an electron moving in a hydrogen atom along a circular orbit of radius `r= 50 p m` would have fallen onto the nucleus. For the sake of simplicity assume the vector `W` to be permanently directed toward the centre of the atom.

Explanation

Moving around the nucleus, the electron raidates and its energy decreases. This means that the electron gets under the nucleus. By the statement of the problem we can assume that the electron is always moving in a circular orbit and the radial acceleration by Newton's law is
`w=(e^(2))/((4 pi epsilon_(0))mr^(2))`
directed inwards. Thus
`(dE)/(dt)= -(mu_(0)e^(6))/(6pi c)(1)/((4pi epsilon_(0))^(2)m^(2)r^(4))` On the other hand in a ciruclar orbit
`E= -(e^(2))/((4pi epsilon_(0))2r)`
So `(e^(2))/((4pi epsilon_(0))2r^(2))(dr)/(dt)= -(mu e^(6))/((4pi epsilon_(0))6pi cm^(2)r^(4))`
or `(dr)/(dt)= -(mue^(4))/((4pi epsilon_(0))3 picm^(2)r^(2))`
Intergrating
`r^(3)=r_(0)^(3)-(mu_(0)e^(4))/(4pi^(2)epsilon_(0)cm^(2))t`
and the radius falls to zero in
`t_(0)=(4pi^(2)epsilon_(0)cm^(2)r_(0)^(3))/(mu_(0)e^(4))sec.= 13.1ps`