In the Bohr model of hydrogen,the force on th | vedclass

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(B) The electrostatic force $F$ between the nucleus and the electron is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$,which

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$F \propto \frac{1}{n^4}$

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(B) The electrostatic force $F$ between the nucleus and the electron is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$,which implies $F \propto \frac{1}{r^2}$.According to the Bohr model,the radius of the $n^{th}$ orbit is given by $r_n = a_0 n^2$,which implies $r \propto n^2$.Substituting the relation $r \propto n^2$ into the force equation,we get $F \propto \frac{1}{(n^2)^2} = \frac{1}{n^4}$.Therefore,the force on the electron is inversely proportional to the fourth power of the principal quantum number $n$.

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

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