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.Thus,the ratio of radii is $\frac{r_1}{r_2} = \lef

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.Thus,the ratio of radii is $\frac{r_1}{r_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.Given $\frac{r_1}{r_2} = \frac{3}{5}$ and $A_1 = 27$,we have $\frac{3}{5} = \left(\frac{27}{A_2}\right)^{1/3}$.Cubing both sides,we get $\frac{27}{125} = \frac{27}{A_2}$,which implies $A_2 = 125$.The number of neutrons $N$ is given by $N = A - Z$,where $Z$ is the atomic number.For nucleus $X$,$A = 125$ and $Z = 52$.Therefore,$N = 125 - 52 = 73$.

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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