When a deuterium is bombarded on $$_8{O^{16}}$$  nucleus, an $$\alpha $$-particle is emitted, then the product nucleus is

Solution :
Let the unknown product nucleus be $$_Z{X^A}.$$
The reaction can be written as
\[\begin{array}{*{20}{c}} {_8{O^{16}}}\\ {\left( {{\rm{Oxygen}}} \right)} \end{array} + \begin{array}{*{20}{c}} {_1{H^2}}\\ {\left( {{\rm{deuterium}}} \right)} \end{array} \to \begin{array}{*{20}{c}} {_Z{X^A}}\\ {\left( {{\rm{unknown}}\,{\rm{nucleus}}} \right)} \end{array} + \begin{array}{*{20}{c}} {_2H{e^4}}\\ {\left( {\alpha - {\rm{particle}}} \right)} \end{array}\]
Conservation of mass number between product and reactant of above reaction gives,
$$16 + 2 = A + 4 \Rightarrow A = 14$$
Conservation of atomic number between reactant and product of above reaction gives
$$8 + 1 = Z + 2 \Rightarrow Z = 7$$
Thus, the unknown product nucleus is nitrogen $$\left( {_7{N^{14}}} \right).$$
NOTE
Fusion reaction can take place at very high temperature $$\left( { \approx {{10}^8}K} \right)$$   and very high pressure which can be provided at sun or by fission of atom bomb.