Question

The most common kind of iron nucleus has a mass number of $56$. Find the radius of the nucleus. $(\text{Take}{R}_0=1.2\times {10}^{-15}\text{m}:(56{)}^{\frac{1}{3}}=3.82)$

• A. $1.2\text{fm}$
• B. $1.8\text{fm}$
• C. $4.6\text{fm}$
• D. $5.2\text{fm}$

Answer 1

The correct option is C $4.6\text{fm}$

The shape of the nucleus is approximately spherical, and its radius is approximately related to the mass number as

$R=R_0{A}^{\frac{1}{3}}$

$\Rightarrow R=(1.2\times {10}^{-15}\text{m})(56{)}^{\frac{1}{3}}$

$\Rightarrow R=4.6\times {10}^{-15}\text{m}=4.6\text{fm}$

Hence, option $\left(C\right)$ is correct.

Why this Question? The shape of the nucleus is approximately spherical, and its radius is approximately related to the mass number by $R=R_0{A}^{\frac{1}{3}}$