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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2024
  • Hard
  • Numerical

The radius of a nucleus of mass number \(64\) is \(4.8\) fermi. Then the mass number of another nucleus having radius of \(4\) fermi is \(\frac{1000}{x}\), where \(x\) is _______.

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Answer & Solution

Correct Answer

(A) \(27\)

Step-by-step Solution

Detailed explanation

\(\mathrm{R}=\mathrm{R}_0 \mathrm{~A}^{1 / 3}\)

\(\mathrm{R}^3 \propto \mathrm{A}\)

\(\left(\frac{4.8}{4}\right)^3=\frac{64}{\mathrm{~A}}\)

\(=\frac{64}{\mathrm{~A}}=(1.2)^3\)

\(\frac{64}{\mathrm{~A}}=1.44 \times 1.2\)

\(\mathrm{~A}=\frac{64}{1.44 \times 1.2}=\frac{1000}{\mathrm{x}}\)

\(\mathrm{x}=\frac{144 \times 12}{64}=27\)

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