Q162 / 180 · KCET 2024 · Physics

The ratio of volume of $\mathrm{Al}^{27}$ nucleus to its surfactarea is (Given, $R_0=1.2 \times 10^{-15} \mathrm{~m}$ )

  • A
    $2.1 \times 10^{-15} \mathrm{~m}$
  • B
    $1.3 \times 10^{-15} \mathrm{~m}$
  • C
    $0.22 \times 10^{-15} \mathrm{~m}$
  • Correct answer
    D
    $1.2 \times 10^{-15} \mathrm{~m}$

Solution

$\frac{\text { Volume of } \mathrm{Al}^{27} \text { nucleus }}{\text { Surface area }}=\frac{\frac{4}{3} \pi R^3}{4 \pi R^2}=\frac{R}{3}$ $=\frac{1}{3}\left(R_0 A^{\frac{1}{3}}\right) \quad\left[\because R=R_0 A^{\frac{1}{3}}\right]$ $=\frac{1}{3} \times 1.2 \times 10^{-15}(27)^{\frac{1}{3}}$ $=1.2 \times 10^{-15} \mathrm{~m}$

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

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