KCET · Physics · Nuclear Physics
The radius of \({ }_{29} \mathrm{Cu}^{64}\) nucleus in Fermi is (given \(\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}\) )
- A \(9.6\)
- B \(4.8\)
- C \(1.2\)
- D \(7.7\)
Answer & Solution
Correct Answer
(B) \(4.8\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3} &=1.2 \times 10^{-15} \times(64)^{1 / 3} \\ &=4.8 \times 10^{-15} \mathrm{~m} \end{aligned}\) or \(\quad \mathrm{R}=4.8\) fermi
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