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(B) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number.Given the ratio of the radius of the target nucleus

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$26$

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(B) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number.Given the ratio of the radius of the target nucleus $(R)$ to the radius of the Helium nucleus $(R_{He})$ is $14^{1/3}$.Using the relation $\frac{R}{R_{He}} = \left( \frac{A}{A_{He}} \right)^{1/3}$,we have $14^{1/3} = \left( \frac{A}{4} \right)^{1/3}$.Cubing both sides,we get $14 = \frac{A}{4}$,which implies $A = 56$.The mass number $A$ is the sum of protons $(Z)$ and neutrons $(N)$.Given $N = 30$,we have $Z + 30 = 56$.Therefore,$Z = 56 - 30 = 26$.

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

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