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Q.
Two nuclei have their mass numbers in the ratio of $1 : 3$. The ratio of their nuclear densities would be
AIPMTAIPMT 2008Atoms
Solution:
Step 1: Concept used:
Their radius of any nucleus is given by:
$R=R_{0} A^{1 / 3}$
Here $R_{0}$ is the constant and $A$ is the atomic mass number.
$A _{1}: A _{2}=1: 3 \text { (given) }$
Density is expressed as the ratio of mass to volume.
Since mass number is directely proportinal to mass then we can write density as:
$\rho=\frac{ A }{\frac{4}{3} \pi R ^{3}}$
$\rho \alpha \frac{ A }{ R ^{3}}$
$\rho \alpha \frac{ A }{\left( R _{0} A ^{1 / 3}\right)^{3}} $
$\rho \alpha \frac{1}{ R _{0}{ }^{3}}$
Density of nuclear matter is independent of mass number, so the required ratio is $1: 1$ Their nuclear densities will be the same.