Q142 / 200 · NEET 2012 (Screening) · Physics

If the nuclear radius of ${ }^{27} \mathrm{Al}$ is $3.6 \mathrm{Fermi}$, the approximate nuclear radius of ${ }^{64} \mathrm{Cu}$ in Fermi is

  • A
    2.4
  • B
    1.2
  • Correct answer
    C
    4.8
  • D
    3.6

Solution

Nuclear radius $r \propto A^{1 / 3}$, where $A$ is mass number $\begin{aligned} & r=r_0 A^{1 / 3} \\ & r=r_0(27)^{1 / 3}=3 r_0 \\ & r_0=\frac{3.6}{3}=1.2 \mathrm{fm} \end{aligned}$ For ${ }^{64} \mathrm{Cu}$ $\begin{aligned} r & =r_0 A^{1 / 3} \\ & =1.2 \mathrm{fm}(64)^{1 / 3} \\ & =4.8 \mathrm{fm} \end{aligned}$

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

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