According to Bohr's theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the `n^(th)` orbit is proportional to :
(n = principal quantum number)

Options

A. `(1)/(n^(3))`
B. `(1)/(n^(5))`
C. `n^(5)`
D. `n^(2)`

Answer

B

Explanation

Electric current due to electron motion in `n^("th")` orbit is given by
`I_(n)=e upsilon_(n)`
For hydrogen atom,
Frequency of electron in `n^("th")` orbit is `upsilon_(n)=((1)/(4piepsilon_(0)))^(2)(4pi^(2)e^(4)m)/(n^(3)h^(3))" ….(i)"`
Radius of `n^("th")` orbit is `r_(n)=(4piepsilon_(0)n^(2)h^(2))/(4pi^(2)e^(2)m)" ....(ii)"`
`I_(n)=((1)/(4piepsilon_(0)))^(2)(4piepsilon_(0))^(2)(4pi^(2)e^(5)m)/(n^(3)h^(3))" (Using (i)) ....(iii)"`
The magnetic field at the centre of the current carrying coil is given by `B=(mu_(0)I)/(2R)`
Hence, the magnetic field produced at the nucleus of a hydrogen atoms due to electron motion in `n^("th")` orbit is
`B_(n)=(mu_(0)I_(n))/(2r_(n))=((1)/(4piepsilon_(0)))^(3)(mu_(0)8pi^(4)e^(7)m^(2))/(n^(5)h^(5))" (Using (ii) and (iii))"`
`therefore B prop (1)/(n^(5))`