Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

If the nuclear radius of $^{27}\text{Al}$ is 3.6 Fermi, the approximate nuclear radius of $^{64}\text{Cu}$ in Fermi is:

2.4

1.2

4.8

3.6

Formula: The nuclear radius ( $R$ ) is related to the mass number ( $A$ ) by the empirical formula: $R = R_0 A^{1/3}$ where $R_0$ is a constant ( $1.2 \text{ fm}$ ).
Set up Ratio: comparing the radii of Copper (Cu) and Aluminum (Al): $\frac{R_{\text{Cu}}}{R_{\text{Al}}} = \frac{R_0 (A_{\text{Cu}})^{1/3}}{R_0 (A_{\text{Al}})^{1/3}} = \left( \frac{A_{\text{Cu}}}{A_{\text{Al}}} \right)^{1/3}$
Substitution: Given $A_{\text{Al}} = 27$ , $R_{\text{Al}} = 3.6 \text{ fm}$ , and $A_{\text{Cu}} = 64$ : $\frac{R_{\text{Cu}}}{3.6} = \left( \frac{64}{27} \right)^{1/3}$
Calculation: $\left( \frac{64}{27} \right)^{1/3} = \frac{(4^3)^{1/3}}{(3^3)^{1/3}} = \frac{4}{3}$ $R_{\text{Cu}} = 3.6 \times \frac{4}{3} = 1.2 \times 4 = 4.8 \text{ fm}$
Conclusion: The nuclear radius of $^{64}\text{Cu}$ is approximately 4.8 Fermi.

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