Antiparticle of electron is

Two nuclei have their mass numbers in the ratio of $1 : 3.$ The ratio of their nuclear densities would be 

An alpha nucleus of energy $\frac{1}{2}m{v^2}$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason $R$. Assertion A : The nuclear density of nuclides ${ }_5^{10} B ,{ }_3^6 Li ,{ }_{26}^{56} Fe ,{ }_{10}^{20} Ne$ and ${ }_{83}^{209} Bi$ can be arranged as $\rho_{ Bi }^{ N }>\rho_{ Fe }^{ N }>\rho_{ Ne }^{ N }>\rho_{ B }^{ N }>\rho_{ Li }^{ N }$.

Reason $R$ : The radius $R$ of nucleus is related to its mass number $A$ as $R=R_0 A^{1 / 3}$, where $R_0$ is a constant.

In the light of the above statement, choose the correct answer from the options given below :

Which of the following is true for a sample of isotope containing ${U^{235}}$ and ${U^{238}}$

A reaction between a proton and $_8{O^{18}}$ that produces $_9{F^{18}}$ must also liberate