The ratio of the radii of nuclei _{13}Al^{27} | vedclass

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(B) The radius of a nucleus is given by the relation $r = R_0 A^{1/3}$,where $A$ is the mass number.Therefore,the ratio of the radii is $\frac{r_1}{r_

Vedclass · Accepted Answer

$73$

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(B) The radius of a nucleus is given by the relation $r = R_0 A^{1/3}$,where $A$ is the mass number.Therefore,the ratio of the radii is $\frac{r_1}{r_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$.Given $\frac{r_1}{r_2} = \frac{3}{5}$,$A_1 = 27$,and $A_2 = A$.Substituting the values: $\frac{3}{5} = \left( \frac{27}{A} \right)^{1/3}$.Cubing both sides: $\frac{27}{125} = \frac{27}{A}$.This gives $A = 125$.The number of neutrons $N$ is given by $N = A - Z$,where $Z$ is the atomic number.For nucleus $X$,$Z = 52$ and $A = 125$.Thus,$N = 125 - 52 = 73$.

The ratio of the radii of nuclei $_{13}Al^{27}$ and $_{52}X^A$ is $3:5$. The number of neutrons in the nucleus of $X$ will be:

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