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GUJCET · Physics · Nuclei
If the radii of \({ }_{13}^{27} \mathrm{~A} l\) and \({ }_{30}^{64} \mathrm{Zn}\) in nucleus are \(R_{1}\) and \(R_{2}\) respectively then \(\frac{R_{1}}{R_{2}}=\) \(\qquad\)
- A \(\frac{64}{27}\)
- B \(\frac{3}{4}\)
- C \(\frac{27}{64}\)
- D \(\frac{4}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
B.\(\frac{3}{4}\)
from \(R=R_{0} A^{\frac{1}{3}}\)
\[
\begin{aligned}
\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{64}\right)^{\frac{1}{3}} \\
\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{3^{3}}{4^{3}}\right)^{\frac{1}{3}} \\
\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{4}
\end{aligned}
\]
from \(R=R_{0} A^{\frac{1}{3}}\)
\[
\begin{aligned}
\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{64}\right)^{\frac{1}{3}} \\
\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{3^{3}}{4^{3}}\right)^{\frac{1}{3}} \\
\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{4}
\end{aligned}
\]
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