The order of magnitude of the density of a ur | vedclass

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(B) The mass of a nucleus with mass number $A$ is approximately $m = A 	imes m_p$,where $m_p$ is the mass of a proton.The radius of a nucleus is give

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${10^{17}} \ kg/m^3$

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(B) The mass of a nucleus with mass number $A$ is approximately $m = A 	imes m_p$,where $m_p$ is the mass of a proton.The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0 \approx 1.2 	imes 10^{-15} \ m$.The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.The density $ho$ is given by $ho = \frac{m}{V} = \frac{A 	imes m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{m_p}{\frac{4}{3} \pi R_0^3}$.Substituting the values: $ho = \frac{1.67 	imes 10^{-27}}{\frac{4}{3} 	imes 3.14 	imes (1.2 	imes 10^{-15})^3} \approx \frac{1.67 	imes 10^{-27}}{7.24 	imes 10^{-45}} \approx 2.3 	imes 10^{17} \ kg/m^3$.Thus,the order of magnitude is $10^{17} \ kg/m^3$.

The order of magnitude of the density of a uranium nucleus is: (Given: ${m_p} = 1.67 \times 10^{-27} \ kg$)

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