ExamBro
ExamBro
KCET · Physics · Nuclear Physics

The radius of \({ }_{29} \mathrm{Cu}^{64}\) nucleus in Fermi is (given \(\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}\) )

  1. A \(9.6\)
  2. B \(4.8\)
  3. C \(1.2\)
  4. D \(7.7\)
Verified Solution

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
Same subject
Explore more questions on app