Question

If the radius of a nucleus ${}^{256}X$ is 8 fermi, then the radius of ${}^{4}He$ nucleus will be:

• A. $16$ fermi
• B. $2$ fermi
• C. $32$ fermi
• D. $4$ fermi

Answer 1

Answer: (B)

$2$ fermi

$R=R_0{\left(A\right)}^{1/3}$

$\therefore$ $\frac{R_1}{R_2}={\left(\frac{A_1}{A_2}\right)}^{1/3}={\left(\frac{256}{4}\right)}^{1/3}=4$

$R_2=\frac{R_1}{4}=2$ fermi

Hence, the correct option is $\text{B}$