If the radius of ${}^{27}_{13}\mathrm{Al}$ nucleus is taken to be $R_{\mathrm{Al}}$, then the radius of ${}^{1

Question

If the radius of ${}^{27}_{13}\mathrm{Al}$ nucleus is taken to be $R_{\mathrm{Al}}$, then the radius of ${}^{125}_{53}\mathrm{Te}$ nucleus is near:

Accepted Answer

1. **Formula:** The nuclear radius $R$ is related to the mass number $A$ by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant ($1.1 	imes 10^{-15}$ m) .
2. **Apply to Aluminum:** For Al, mass number $A_1 = 27$.
   $$ R_{\mathrm{Al}} = R_0 (27)^{1/3} = 3R_0 $$
3. **Apply to Tellurium:** For Te, mass number $A_2 = 125$.
   $$ R_{\mathrm{Te}} = R_0 (125)^{1/3} = 5R_0 $$
4. **Calculate Ratio:**
   $$ \frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}} = \frac{5R_0}{3R_0} = \frac{5}{3} $$
   $$ R_{\mathrm{Te}} = \frac{5}{3} R_{\mathrm{Al}} $$

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