Derive an expression for the magnetic field at the site of the nucleus in a hydrogen atom due to the circular motion of the electron Assume that the atom is in its ground state and the answer in terms of fundamental constants

Explanation

For circular motion of electron we can write the following equation :
`1/(4piepsilon_0)e^2/r^2=(mv^2)/r`
`rArr mv^2r=e^2/(4piepsilon_0)`…(i)
Bohr.s quantisation rule can be written as follows for the first orbit :
`mvr=h/(2pi)`…(ii)
On dividing equation (ii) from equation (i) we get the following
`rArr (mv^2r)/(mvr)=((e^2)/(4piepsilon_0))/(h/(2pi))`
`rArr v=e^2/(2epsilon_0h)`....(iii)
Substituting v from equation (iii) into equation (ii) we get the following :
`r=(epsilon_0h^2)/(pime^2)`...(iv)
Electron crosses any point on its path after every `2pir//v` time hence equivalent current due to its motion can be written as follows :
`i=e/((2pir)/v)=(ev)/(2pir)`
Magnetic field at the centre of circular current can be written as follows :
`B=(mu_0i)/(2r)=(mu_0ev)/(4pir^2)`
Substituting values from equation (iii) and (iv) we get the following :
`B=(mu_0e(e^2)/(2epsilon_0h))/(4pi((epsilon_0h^2)/(pime^2))^2)`
`B=(mu_0e^7pim^2)/(8epsilon_0^3h^5)`