Derive an expression for the magnetic field at the site of the necleas 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 lerms of fandmental constants

Explanation

We have
`mv^(2))/(r ) = (v^(2))/(4 pi epsilon_(0)r^(2))`
or, `v6(2)r= (e^(2))/(4 pi epsilon_(0)m)`…(i)
From bohr's quantination rule in ground state
`ur= (h)/(2 pim)`..(ii)
From (i) and(ii)
`u = *e^(2))/(2 epsilon_(0)h)`…(iii)
and` r = (epsilobn_(0)h^(2))/(pi me^(2))`...(iv)
As the electron moves along a circle time any point on the circle `(e)/(2 pir)` times per unit time The charge crossing per unit time is the current is `B = (mu_(0)i)/(4 pi r^(2))`
`From (iii) and (iv)
`B = (mu_(0)e)/(4 pi) (e6(2))/(2 epsilon_(0)h) xx (pi^(2)m^(2)e^(2))/(epsilos_(0)^(2)h^(4))`
`= (mu_(0)e^(2)pi m6(2))/(8 epsilon_(0)^(2)h^(2))`