Question

Highly energetic electrons are bombarded on a target of an element containing $30$ neutrons. The ratio of radii of nucleus to that of Helium nucleus is ${\left(14\right)}^{1/3}$. The atomic number of nucleus will be

• A. $25$
• B. $26$
• C. $56$
• D. $30$

Answer 1

The correct option is B $26$

Using $R=R_0{A}^{1/3}$
$\frac{R_1}{R_2}={\left(\frac{A_1}{A_2}\right)}^{1/3}$
$\frac{R}{R_{He}}={\left(\frac{A}{4}\right)}^{\frac{1}{3}}$
${\left(14\right)}^{1/3}={\left(\frac{A}{4}\right)}^{\frac{1}{3}}\Rightarrow A=56$
So, $=56-30=26$