Question

If the ratio of the radius of a nucleus with $61$ neutrons to that of helium nucleus is $3$, then the atomic number of this nucleus is:

• A. $27$
• B. $47$
• C. $51$
• D. $61$
• E. $108$

Answer 1

The correct option is B $47$

The radius is directly proportional to $\frac{1}{3}$ power of mass number, i.e, $R\propto A^{1/3}$
or $\frac{A_1}{A_2}={\left(\frac{R_1}{R_2}\right)}^3={\left(\frac{1}{3}\right)}^3=\frac{1}{27}$
$\Rightarrow \frac{A_2}{A_1}=27$
$\Rightarrow A_2=A_1\times 27$
$=4\times 27=108$
$\Rightarrow Z_2=A_2-61$
$=108-61=47$.