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(C) According to Bohr's theory for an electron in a circular orbit,the centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \fr

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$1 : 8$

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(C) According to Bohr's theory for an electron in a circular orbit,the centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{ke^2}{r^2}$,which implies $v \propto r^{-1/2}$.The time period $T$ is given by $T = \frac{2\pi r}{v}$.Substituting $v \propto r^{-1/2}$,we get $T \propto \frac{r}{r^{-1/2}} = r^{3/2}$.Given radii are $r_1 = R$ and $r_2 = 4R$.Therefore,the ratio of time periods is $\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} = \left(\frac{R}{4R}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$.Thus,the ratio is $1 : 8$.

In an atom,two electrons move around the nucleus in circular orbits of radii $R$ and $4R$. The ratio of the time taken by them to complete one revolution is:

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