Given the mass of an iron nucleus as 55.85 , | vedclass

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Question

(A) The mass of the iron nucleus is $M = 55.85 \, u$. Converting this to kilograms: $M = 55.85 	imes 1.66 	imes 10^{-27} \, kg \approx 9.27 	imes 1

Vedclass · Accepted Answer

$2.29 	imes 10^{17} \, kg \, m^{-3}$

Vedclass · Answer

(A) The mass of the iron nucleus is $M = 55.85 \, u$. Converting this to kilograms: $M = 55.85 	imes 1.66 	imes 10^{-27} \, kg \approx 9.27 	imes 10^{-26} \, kg$.The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0 = 1.2 	imes 10^{-15} \, m$.The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.Nuclear density $ho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R_0^3 A}$.Substituting the values: $ho = \frac{55.85 	imes 1.66 	imes 10^{-27}}{\frac{4}{3} \pi (1.2 	imes 10^{-15})^3 	imes 56}$.Since $M \approx A 	imes m_p$ (where $m_p$ is the mass of a nucleon),the density simplifies to $ho = \frac{m_p}{\frac{4}{3} \pi R_0^3} \approx 2.29 	imes 10^{17} \, kg \, m^{-3}$.

Given the mass of an iron nucleus as $55.85 \, u$ and $A=56$,find the nuclear density.

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