If the nuclear radius of ${}^{27}\mathrm{Al}$ is 3.6 Fermi, the approximate nuclear radius of ${}^{64}\mathrm{

Question

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

Accepted Answer

1. **Formula:** The nuclear radius $R$ is related to the mass number $A$ by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant .
2. **Identify Values:**
   - For Aluminum (${}^{27}\mathrm{Al}$): $A_1 = 27$, $R_1 = 3.6$ Fermi.
   - For Copper (${}^{64}\mathrm{Cu}$): $A_2 = 64$, $R_2 = ?$
3. **Calculation:**
   Taking the ratio of the radii:
   $$ \frac{R_{Cu}}{R_{Al}} = \left( \frac{A_{Cu}}{A_{Al}} ight)^{1/3} $$
   $$ \frac{R_{Cu}}{3.6} = \left( \frac{64}{27} ight)^{1/3} $$
   $$ \frac{R_{Cu}}{3.6} = \frac{(4^3)^{1/3}}{(3^3)^{1/3}} = \frac{4}{3} $$
   $$ R_{Cu} = 3.6 	imes \frac{4}{3} = 1.2 	imes 4 = 4.8 	ext{ Fermi} $$

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