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(D) According to Kepler's Third Law of planetary motion,which also applies to electrons in circular orbits under a central force,the square of the tim

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$1/8$

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(D) According to Kepler's Third Law of planetary motion,which also applies to electrons in circular orbits under a central force,the square of the time period $T$ is proportional to the cube of the radius $R$ of the orbit: $T^{2} \propto R^{3}$.Given the radii are $R_{1} = R$ and $R_{2} = 4R$.The ratio of the time periods is given by:$\frac{T_{1}}{T_{2}} = \sqrt{\frac{R_{1}^{3}}{R_{2}^{3}}}$Substituting the values:$\frac{T_{1}}{T_{2}} = \sqrt{\frac{R^{3}}{(4R)^{3}}} = \sqrt{\frac{R^{3}}{64R^{3}}} = \sqrt{\frac{1}{64}} = \frac{1}{8}$.Thus,the ratio of the time taken 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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