1.

Magnetic field at the centre of a Bohr's hypotheticalhydrogen atom in the n^(th) orbit of the electron is

Answer»

directly proportional to charge of electron e
directly proportional to `e^2`
inversely proportional to `n^5`
directly proportional to `n^5`

Solution :An electron revolving in a circular orbit in a hydrogen ATOM is equivalent to a current of `eupsilon`.
Magnetic FIELD at the centre of Bohr.s HYPOTHETICAL orbit
`B=(mu_0I)/(2r_n) = (mu_0 eupsilon)/(2r_n)` ….(i)
where `r_n` is the radius of `n^(TH)` orbit.
The electric force between the nucleus and electron in the `n^(th)` orbit provides the centripetal force for circular motion.
`therefore e^2/(4piepsilon_0r_n^2)=(mv_n^2)/r_n` or `r_n=e^2/(4piepsilon_0mv_n^2)`....(ii)
According to Bohr.s quantisation condition
`mv_n r_n=nħ` or `v_n=(nħ)/(mr_n)` ...(iii)
From equations (ii) and (iii) , we get
`r_n=(4pi epsilon_0 ħ^2 n^2)/(me^2)`
Then equation (i) becomes
`B=(mu_0e)/(2r_n) (v_n/(2pir_n))=(mu_0e)/(4pi) ((nħ//mr_n))/r_n^2`
`=(mu_0enħ)/(4pim)=((me^2)/(4piepsilon_0ħ^2n^2))^3=(mu_0pim^2e^7)/(8epsilon_0^3 h^5 n^5) [ because ħ=h/(2pi)]`
`therefore B prop 1/n^5`


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