If the binding energy per nucleon in $${}_3^7Li$$ and $${}_2^4He$$ nuclei are $$5.60$$ $$MeV$$ and $$7.06$$ $$MeV$$ respectively, then in the reaction $$$p + {}_3^7Li \to 2\,{}_2^4He$$$
energy of proton must be

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

When $${}_3L{i^7}$$ nuclei are bombarded by protons, and the resultant nuclei are $${}_4B{e^8}$$, the emitted particles will be

If radius of the $$\matrix{ {27} \cr {13} \cr } $$ $$Al$$ nucleus is estimated to be $$3.6$$ fermi then the radius of $$\matrix{ {125} \cr {52} \cr } \,Te$$ nucleus is estimated to be nearly