Q142 / 200 · NEET 2007 · Physics

If the nucleus ${ }_{13}^{27} \mathrm{Al}$ has a nuclear radius of about $3.6 \mathrm{fm}$, then ${ }_{32}^{125} \mathrm{Te}$ would have its radius approximately as:

  • A
    $9.6 \mathrm{fm}$
  • B
    $12.0 \mathrm{fm}$
  • C
    $4.8 \mathrm{fm}$
  • Correct answer
    D
    $6.0 \mathrm{fm}$.

Solution

Nuclear radii \(R=\left(R_0\right) A^{1 / 3}\) where, \(A\) is the mass number. \(\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}}=\left(\frac{A_{\mathrm{Te}}}{A_{\mathrm{Al}}}\right)^{1 / 3}=\left(\frac{125}{27}\right)^{1 / 3}=\left(\frac{5}{3}\right)\) or, \(\quad R_{\mathrm{Te}}=\frac{5}{3} \times R_{\mathrm{Al}}=\frac{5}{3} \times 3.6=6 \mathrm{fm}\). (Given \(R_{\mathrm{Al}}=3.6 \mathrm{fm}\))

Hence, the correct answer is option: 'D'.

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