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 _______.
Pick an option to check your answer.
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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