Question

A deuteron is bombarded on ⁸O₁₆ nucleus then the α-particle is emitted then the product nucleus is

• A. ⁷N₁₃
• B. ⁵B₁₀
• C. ⁴Be₉
• D. ⁷N₁₄

Answer 1

The Correct Option is D

To solve the problem, we need to understand the nuclear reaction in which a deuteron (^2\text{H}) is bombarded on an ^{16}\text{O} nucleus, resulting in the emission of an α-particle (helium nucleus, ^4\text{He}). We need to determine the resultant nucleus.

The nuclear reaction can be represented as:

^2\text{H} + ^{16}\text{O} \rightarrow \,\text{Product Nucleus} + \,^4\text{He}

Let's analyze the reaction step-by-step:

1. Consider the reactants: The deuteron has a mass number of 2 and an atomic number of 1, denoted by ^2\text{H}_1. The ^{16}\text{O} nucleus has a mass number of 16 and an atomic number of 8, denoted by ^{16}\text{O}_8.
2. Balance the atomic numbers (protons):
   • Total atomic number on the left side = 1 (from deuteron) + 8 (from oxygen) = 9.
   • After emitting an α-particle (having atomic number 2), the remaining atomic number = 9 - 2 = 7.
3. Balance the mass numbers (nucleons):
   • Total mass number on the left side = 2 (from deuteron) + 16 (from oxygen) = 18.
   • After emitting an α-particle (with mass number 4), the remaining mass number = 18 - 4 = 14.
4. Thus, the resultant nucleus must have a mass number of 14 and an atomic number of 7, which corresponds to ^7\text{N}_{14}.

Therefore, the product nucleus is ^7\text{N}_{14}.

Hence, the correct option is ^7\text{N}_{14}.