Question:
Order of magnitude of density of uranium nucleus is \((m_p = 1.6 \times 10^{-27}\) kg\()\)
Order of magnitude of density of uranium nucleus is \((m_p = 1.6 \times 10^{-27}\) kg\()\)
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Nuclear density is approximately the same for all nuclei ($\sim 2\times10^{17}$ kg/m$^3$) because both mass and volume scale as $A$.
Updated On: Apr 20, 2026
- \(10^{20}\) kg/m\(^3\)
- \(10^{17}\) kg/m\(^3\)
- \(10^{14}\) kg/m\(^3\)
- \(10^{11}\) kg/m\(^3\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Nuclear radius follows \(R = R_0 A^{1/3}\) where \(R_0 = 1.25 \times 10^{-15}\) m, and mass \(\approx A m_p\).
Step 2: Detailed Explanation:
\[ \rho = \frac{m}{V} = \frac{A m_p}{\frac{4}{3}\pi R_0^3 A} = \frac{3m_p}{4\pi R_0^3} \] \[ = \frac{3 \times 1.67\times10^{-27}}{4\pi (1.25\times10^{-15})^3} \approx 2 \times 10^{17} \text{ kg/m}^3 \] This is independent of \(A\) (all nuclei have approximately the same density).
Step 3: Final Answer:
Nuclear density \(\approx 10^{17}\) kg/m\(^3\).
Nuclear radius follows \(R = R_0 A^{1/3}\) where \(R_0 = 1.25 \times 10^{-15}\) m, and mass \(\approx A m_p\).
Step 2: Detailed Explanation:
\[ \rho = \frac{m}{V} = \frac{A m_p}{\frac{4}{3}\pi R_0^3 A} = \frac{3m_p}{4\pi R_0^3} \] \[ = \frac{3 \times 1.67\times10^{-27}}{4\pi (1.25\times10^{-15})^3} \approx 2 \times 10^{17} \text{ kg/m}^3 \] This is independent of \(A\) (all nuclei have approximately the same density).
Step 3: Final Answer:
Nuclear density \(\approx 10^{17}\) kg/m\(^3\).
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