Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. ratio of

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13.Nuclei

medium

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

$25$

$26$

$56$

$30$

Solution

(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}}$
==> $A = 56$ so $Z = 56 -30 = 26.$

Standard 12

Physics