Nucleus: Practice Problem & Solution
If the nucleus $^{27}_{13}Al$ has nuclear radius of about 3.6 fm, then $^{125}_{52}Te$ would have its radius approximately as: (2007)
Solution Explained:
To solve this problem, we apply the core principles of Nucleus. Understanding the underlying formula is key to arriving at the correct answer below:
Using the relation $R \propto A^{1/3}$, we get $R_{Te} = R_{Al} (A_{Te} / A_{Al})^{1/3}$. $R_{Te} = 3.6 \times (125 / 27)^{1/3} = 3.6 \times (5 / 3) = 6.0 fm$.
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