If radius of 13Al27 nucleus is estimated to be 3.6 fm, then the radius of 52Te125 nucleus will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6.0 fm
Nuclear density is nearly the same for all nuclei, so the volume of a nucleus is proportional to its mass number A and the radius varies as \( A^{1/3} \).
\( \displaystyle R = R_0 A^{1/3} \)
\( \displaystyle \frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}} = \left(\frac{A_{\mathrm{Te}}}{A_{\mathrm{Al}}}\right)^{1/3} \)
\( \displaystyle \frac{R_{\mathrm{Te}}}{3.6\,\mathrm{fm}} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3} \)
\( \displaystyle R_{\mathrm{Te}} = 3.6\,\mathrm{fm}\times\frac{5}{3} \)
\( \displaystyle R_{\mathrm{Te}} = 6.0\,\mathrm{fm} \)