Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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:

(\frac{53}{13})^{1/3} R_{\mathrm{Al}}

\frac{5}{3} R_{\mathrm{Al}}

\frac{3}{5} R_{\mathrm{Al}}

(\frac{13}{53}) R_{\mathrm{Al}}

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 ( $1.1 \times 10^{-15}$ m) .
Apply to Aluminum: For Al, mass number $A_1 = 27$ . $R_{\mathrm{Al}} = R_0 (27)^{1/3} = 3R_0$
Apply to Tellurium: For Te, mass number $A_2 = 125$ . $R_{\mathrm{Te}} = R_0 (125)^{1/3} = 5R_0$
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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