Question

Two nuclei have their mass numbers in the ratio of $1:3$. The ratio of their nuclear densities would be

• A. $(3{)}^{1/3}:1$
• B. $1:1$
• C. $1:3$
• D. $3:1$

Answer 1

Answer: (B)

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.