CUET · PHYSICS · PYQ PAPER 2023
The ratio of the Nuclear density of the gold isotope \(^{197}_{79}\text{Au}\) and the silver isotope \(^{107}_{47}\text{Ag}\) is :
- A \((\frac{197}{107})^{\frac{1}{3}}\)
- B \((\frac{79}{47})^{\frac{1}{3}}\)
- C \((\frac{197}{107})^{\frac{2}{3}}\)
- D \(1:1\)
Answer & Solution
Correct Answer
(D) \(1:1\)
Step-by-step Solution
Detailed explanation
Nuclear density \(\rho = \frac{\text{Mass}}{\text{Volume}}\). \(\rho \approx \frac{A m_n}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \frac{A m_n}{\frac{4}{3}\pi R_0^3 A} = \frac{m_n}{\frac{4}{3}\pi R_0^3}\). Nuclear density is independent of mass number \(A\). Ratio of nuclear density of…
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