The ratio of mass numbers of two nuclei is 1: | vedclass

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(B) The nuclear density $ho$ is defined as the ratio of the mass of the nucleus to its volume.$ho = \frac{	ext{Mass}}{	ext{Volume}} = \frac{A \c

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$1:1$

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(B) The nuclear density $ho$ is defined as the ratio of the mass of the nucleus to its volume.$ho = \frac{	ext{Mass}}{	ext{Volume}} = \frac{A \cdot m_p}{\frac{4}{3}\pi R^3}$,where $A$ is the mass number,$m_p$ is the mass of a nucleon,and $R$ is the nuclear radius.Since the nuclear radius $R$ is given by $R = R_0 A^{1/3}$,we have $R^3 = R_0^3 A$.Substituting this into the density formula:$ho = \frac{A \cdot m_p}{\frac{4}{3}\pi (R_0^3 A)} = \frac{m_p}{\frac{4}{3}\pi R_0^3}$.This shows that the nuclear density $ho$ is independent of the mass number $A$.Therefore,the ratio of the nuclear densities of two nuclei with any mass numbers is always $1:1$.

The ratio of mass numbers of two nuclei is $1:3$. What is the ratio of their nuclear densities?

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