Question 38: Atoms & Nuclei

Question

$$
	ext { If the nuclear radius of }{ }^{27} \mathrm{Al} 	ext { is } 3.6 	ext { fermi, the nuclear radius of }{ }^{125} \mathrm{Fe} 	ext { is }
$$

Accepted Answer

To determine the nuclear radius of $$^{125}\mathrm{Fe}$$, we need to use the relationship between nuclear radius and mass number. The nuclear radius ($$R$$) of a nucleus is given by the empirical formula:
$$R = R_0 A^{1/3}$$
where:
$$R_0$$ is the proportionality constant (approximately 1.2-1.3 fermi) and $$A$$ is the mass number (number of nucleons).
Given the nuclear radius of $$^{27}\mathrm{Al}$$ (Aluminium-27) is 3.6 fermi, we can use this to find $$R_0$$:
$$R_{^{27}\mathrm{Al}} = R_0 \cdot 27^{1/3}$$
We know:
$$R_{^{27}\mathrm{Al}} = 3.6 	ext{ fermi}$$
Therefore:
$$3.6 = R_0 \cdot 27^{1/3}$$
$$R_0 = \frac{3.6}{27^{1/3}}$$
We can solve for $$R_0$$:
$$27^{1/3} = 3$$
$$R_0 = \frac{3.6}{3} = 1.2 	ext{ fermi}$$
Now, using this $$R_0$$, we find the radius of $$^{125}\mathrm{Fe}$$ (Iron-125):
$$R_{^{125}\mathrm{Fe}} = R_0 \cdot 125^{1/3}$$
Substitute $$R_0$$:
$$R_{^{125}\mathrm{Fe}} = 1.2 \cdot 125^{1/3}$$
We need to find $$125^{1/3}$$:
$$125^{1/3} = 5$$
Therefore:
$$R_{^{125}\mathrm{Fe}} = 1.2 \cdot 5 = 6 	ext{ fermi}$$
Converting fermi to meters (since 1 fermi = $$10^{-15}$$ meters):
$$6 	ext{ fermi} = 6 	imes 10^{-15} 	ext{ meters}$$
The correct option is:
Option C
$$6 	imes 10^{-15} \mathrm{~m}$$

Answer

$$6 	imes 10^{-10} \mathrm{~m}$$

Answer

$$6 	imes 10^{-13} \mathrm{~m}$$

Answer

$$6 	imes 10^{-15} \mathrm{~m}$$

Answer

$$6 	imes 10^{-12} \mathrm{~m}$$

Atoms & Nuclei

\text { If the nuclear radius of }{ }^{27} \mathrm{Al} \text { is } 3.6 \text { fermi, the nuclear radius of }{ }^{125} \mathrm{Fe} \text { is }

Choose the correct answer: