The radius of {}_{52}Te^{125} nucleus is 6 te | vedclass

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(B) The relation between the nuclear radius $(R)$ and the mass number $(A)$ is given by the formula: $R = R_0 A^{1/3}$.Therefore,the ratio of the radi

Vedclass · Accepted Answer

$3.6 	imes 10^{-15} 	ext{ m}$

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(B) The relation between the nuclear radius $(R)$ and the mass number $(A)$ is given by the formula: $R = R_0 A^{1/3}$.Therefore,the ratio of the radii of two nuclei is given by: $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}ight)^{1/3}$.Given: $R_1 = 6 	ext{ fermi}$,$A_1 = 125$,and $A_2 = 27$.Substituting the values into the formula:$\frac{6}{R_2} = \left(\frac{125}{27}ight)^{1/3} = \frac{5}{3}$.Solving for $R_2$:$R_2 = \frac{6 	imes 3}{5} = \frac{18}{5} = 3.6 	ext{ fermi}$.Since $1 	ext{ fermi} = 10^{-15} 	ext{ m}$,we have:$R_2 = 3.6 	imes 10^{-15} 	ext{ m}$.

The radius of ${}_{52}Te^{125}$ nucleus is $6 \text{ fermi}$. The radius of ${}_{13}Al^{27}$ nucleus in meters is

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