Question

If the radius of $_{13}\text{Al}^{27}$ nucleus is $3.6\text{ fm}$, then the number of neutrons in a nucleus of atomic number $29$ and radius $4.8\text{ fm}$ is:

• A. $64$
• B. $35$
• C. $42$
• D. $49$

Hint:

In nucleus-radius problems, first find the mass number using $R\propto A^{1/3}$. If the question asks for neutrons, do not forget the final step: $N=A-Z$.

Answer 1

The Correct Option is B

Concept: The nuclear radius is related to the mass number by \[ R=R_0A^{1/3} \] where $R_0$ is a constant. For two nuclei, \[ \frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1/3} \] Also, \[ N=A-Z \] where $N$ is the number of neutrons and $Z$ is the atomic number.

Step 1: Apply the radius relation For aluminium nucleus, \[ A_1=27,\qquad R_1=3.6\text{ fm} \] For the unknown nucleus, \[ A_2=?,\qquad R_2=4.8\text{ fm} \] Thus, \[ \frac{3.6}{4.8}=\left(\frac{27}{A_2}\right)^{1/3} \] \[ \frac{3}{4}=\left(\frac{27}{A_2}\right)^{1/3} \] Cubing both sides, \[ \left(\frac{3}{4}\right)^3=\frac{27}{A_2} \] \[ \frac{27}{64}=\frac{27}{A_2} \] Therefore, \[ A_2=64 \]

Step 2: Calculate the neutron number Given \[ Z=29 \] Hence, \[ N=A-Z=64-29=35 \] Therefore, \[ \boxed{N=35} \]