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

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

If the ratio of the concentration of electrons to that of holes in a semiconductor is $${7 \over 5}$$ and the ratio of currents is $${7 \over 4},$$ then what is the ratio of their drift velocities?