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

Question

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

Accepted Answer

1.  **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 ($1.2 	ext{ fm}$).
2.  **Set up Ratio:** comparing the radii of Copper (Cu) and Aluminum (Al):
    $$ \frac{R_{	ext{Cu}}}{R_{	ext{Al}}} = \frac{R_0 (A_{	ext{Cu}})^{1/3}}{R_0 (A_{	ext{Al}})^{1/3}} = \left( \frac{A_{	ext{Cu}}}{A_{	ext{Al}}} ight)^{1/3} $$
3.  **Substitution:** Given $A_{	ext{Al}} = 27$, $R_{	ext{Al}} = 3.6 	ext{ fm}$, and $A_{	ext{Cu}} = 64$:
    $$ \frac{R_{	ext{Cu}}}{3.6} = \left( \frac{64}{27} ight)^{1/3} $$
4.  **Calculation:**
    $$ \left( \frac{64}{27} ight)^{1/3} = \frac{(4^3)^{1/3}}{(3^3)^{1/3}} = \frac{4}{3} $$
    $$ R_{	ext{Cu}} = 3.6 	imes \frac{4}{3} = 1.2 	imes 4 = 4.8 	ext{ fm} $$
5.  **Conclusion:** The nuclear radius of $^{64}	ext{Cu}$ is approximately 4.8 Fermi.

Step-by-Step Solution

Practice All PHYSICS Questions