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(B) In the Bohr model,the electrostatic force $F$ between the nucleus and the electron is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{

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$F \propto 1/n^4$

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(B) In the Bohr model,the electrostatic force $F$ between the nucleus and the electron is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$.From Bohr's theory,the radius of the $n^{th}$ orbit is $r_n \propto n^2$.Substituting this into the force equation,we get $F \propto \frac{1}{(n^2)^2} = \frac{1}{n^4}$.Alternatively,using centripetal force $F = \frac{mv^2}{r}$,where $v \propto \frac{1}{n}$ and $r \propto n^2$,we get $F \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4}$.

In the Bohr model of the hydrogen atom,the electrostatic force on the electron depends on the principal quantum number $n$ as:

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