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Wave Optics

Question

What is the nuclear radius of ${Fe}^{125},$ if that of Al²⁷ is 3.6 fermi?

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Solution

Relation between nuclear radius and mass number is as follows,

R \propto A^{1 / 3}

This implies,

\frac{R_{Fe}}{R_{Al}} = {(\frac{A_{Fe}}{A_{Al}})}^{1 / 3} = {(\frac{125}{27})}^{1 / 3} = \frac{5}{3} = 1.67

Therefore,
Nuclear radius,

R_{Fe} = 1.67 \times R_{Al}

= \frac{5}{3} \times 3.6 = 6 fermi .