Q. 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 $14^{1/3}$. The atomic number of nucleus will be
A) 25
B) 26
C) 56
D) 30
Explanation
(b) By using $R = R_{0}\ A^{1/3} \Rightarrow \frac{R_{1}}{R_{2}} = \left( \frac{A_{1}}{A_{2}} \right)^{1/3}$
$\Rightarrow \frac{R}{R_{He}} = \left( \frac{A}{4} \right)^{1/3} \Rightarrow (14)^{1/3} = \left( \frac{A}{4} \right)^{1/3}$
$\Rightarrow A = 56$ so $Z = 56 - 30 = 26$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.