MHT CET · Physics · Nuclear Physics
Two nuclei have their mass numbers in the ratio of \(1: 3\). The ratio of their nuclear densities would be
- A \(1: 3\)
- B \(3: 1\)
- C \((3)^{1 / 3}: 1\)
- D \(1: 1\)
Answer & Solution
Correct Answer
(D) \(1: 1\)
Step-by-step Solution
Detailed explanation
Density of nuclear matter is independent of mass number, so the required ratio is \(1: 1\).
Alternative
\(A_{1}: A_{2}=1: 3\)
Their radii will be in the ratio
\(R_{0} A_{1}^{1 / 3}: R_{0} A_{2}^{1 / 3}=1: 3^{1 / 3} \)
\( \text {Density }=\frac{A}{\frac{4}{3} \pi R^{3}} \)
\( \therefore \rho_{A_{1}}: \rho_{A_{2}}=\frac{1}{\frac{4}{3} \pi R_{0}^{3} \cdot 1^{3}}=\frac{3}{\frac{4}{3} \pi R_{0}^{3}\left(3^{1 / 3}\right)^{3}}\)
Their nuclear densities will be the same.
Alternative
\(A_{1}: A_{2}=1: 3\)
Their radii will be in the ratio
\(R_{0} A_{1}^{1 / 3}: R_{0} A_{2}^{1 / 3}=1: 3^{1 / 3} \)
\( \text {Density }=\frac{A}{\frac{4}{3} \pi R^{3}} \)
\( \therefore \rho_{A_{1}}: \rho_{A_{2}}=\frac{1}{\frac{4}{3} \pi R_{0}^{3} \cdot 1^{3}}=\frac{3}{\frac{4}{3} \pi R_{0}^{3}\left(3^{1 / 3}\right)^{3}}\)
Their nuclear densities will be the same.
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