Estimate the dimensions of a hydrogen atom in the nonexcited state, regarding it as an oscillator and assuming the zero-point energy of oscillations to be equal to the kinetic energy of the electron on the first orbit.

Answer

`r=4piepsi_(0)h^(2)//e^(2)m_(e)`

Explanation

The kinetic energy of an orbiting electron is `K=1/2m_(e)v^(2)=1/2m_(e)omega^(2)r^(2)`, the zero-point energy is `epsi_(0)=homega//2`. The natural frequency is `omega=sqrt(k_(el)//m_(e))=sqrt(F//m_(e)A)`, where F is a quasi-elastic force and A the amplitude of the oscillator. Assuming the quasi-elastic

force to be of the Coulomb type and the amplitude to be equal to the radius, we obtain
`omega=sqrt(e^(2)/(4piepsi_(0)m_(e)r^(3)))`
Putting `K=epsi_(0)` and substituting the circular frequency, we obtain after some simple transformations
`r=(4piepsi_(0)h)/(e^(2)m_(e))`
Despite certain arbitrary assumptions, we have obtained a correct expression for the first Bohr radius.