The density of neutrons is of the order

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(D) The density of a neutron is calculated by dividing its mass by its volume. Given the mass of a neutron is approximately $1.67 	imes 10^{-27} \ kg

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$10^{11} \ kg/cc$

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(D) The density of a neutron is calculated by dividing its mass by its volume. Given the mass of a neutron is approximately $1.67 	imes 10^{-27} \ kg$ and its radius is approximately $10^{-15} \ m$. Using the volume formula for a sphere $V = \frac{4}{3} \pi r^3$,the volume is approximately $4.18 	imes 10^{-45} \ m^3$. The density is $ho = \frac{m}{V} \approx 4 	imes 10^{17} \ kg/m^3$. Converting $m^3$ to $cc$ $(1 \ m^3 = 10^6 \ cc)$,the density becomes $4 	imes 10^{11} \ kg/cc$. Thus,the order of magnitude is $10^{11} \ kg/cc$.

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