Question

 

Complete these nuclear reaction.

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Answer 1

Concepts and reason

The concept used to solve this problem is based on nuclear reactions.

In nuclear reactions, the invariant mass must balance for each side of the equation, and the transformation of particles should follow certain conversion laws such as, conversion of charge and mass.

Fundamentals

The nuclear reactions are the reactions in which two nuclear particles or nuclei collide and produce changed product from the initial particles.

Part 1

In the mentioned reaction, the mass number on reactant side is 2. So, the sum of mass number on products side should be equal to 2. Thus, the unknown particle has a mass of $+ 1$ .

Now, the atomic number on reactant side is 1. So, the sum of atomic numbers on products side should be equal to 1. Thus, the unknown particle has an atomic number of 0.

The unknown particle is neutron.

Part 2

In the mentioned reaction, the mass number on reactant side is 61. So, the sum of mass number on products side should be equal to 61. Thus, the unknown particle has a mass of0.

Now, the atomic number on reactant side is 29. So, the sum of atomic numbers on products side should be equal to 29. Thus, the unknown particle has an atomic number of $- 1$ .

The unknown particle is beta particle.

Part 3

In the mentioned reaction, the mass number on reactant side is 210. So, the sum of mass number on products side should be equal to 210. Thus, the unknown particle has a mass of $+ 4$ .

Now, the atomic number on reactant side is 84. So, the sum of atomic numbers on products side should be equal to 84. Thus, the unknown particle has an atomic number of 2.

The unknown particle is alpha particle.

Ans: Part 1

The complete nuclear reaction is written as follows:

$_1^2{\rm{H}} \to _1^1{\rm{H}} + {\rm{n}}$

Part 2

The complete nuclear reaction is written as follows:

$_{29}^{61}{\rm{Cu}} \to _{30}^{61}{\rm{Zn}} + \beta$

Part 3

The complete nuclear reaction is written as follows:

$_{84}^{210}{\rm{Po}} \to _{82}^{206}{\rm{Pb}} + \alpha$