KCET · Physics · Atomic 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 \(0.22 \times 10^{-15} \mathrm{~m}\)
- D \(1.2 \times 10^{-15} \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(1.2 \times 10^{-15} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\frac{\text { Volume of } \mathrm{Al}^{27} \text { nucleus }}{\text { Surface area }}=\frac{\frac{4}{3} \pi R^3}{4 \pi R^2}=\frac{R}{3}\)
\(=\frac{1}{3}\left(R_0 A^{\frac{1}{3}}\right) \quad\left[\because R=R_0 A^{\frac{1}{3}}\right]\)
\(=\frac{1}{3} \times 1.2 \times 10^{-15}(27)^{\frac{1}{3}}\)
\(=1.2 \times 10^{-15} \mathrm{~m}\)
\(=\frac{1}{3}\left(R_0 A^{\frac{1}{3}}\right) \quad\left[\because R=R_0 A^{\frac{1}{3}}\right]\)
\(=\frac{1}{3} \times 1.2 \times 10^{-15}(27)^{\frac{1}{3}}\)
\(=1.2 \times 10^{-15} \mathrm{~m}\)
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