Artificial Radioactivity Fission and Fusion

Artificial Radioactivity Fission and Fusion

Two artificial nuclear reactions were done by Rutherford. In this reactions, Rutherford use α decay to convert ₄¹⁷N to ₈¹⁷O.

₄¹⁷N + ₂⁴He → ₈¹⁷O + ₁¹H

₈¹⁷O is not radioactive element.First artificial radioactive nucleus is ₁₅³⁰P and it is produced by alpha decay of ₁₃²⁷Al.

₁₃²⁷Al + ₂⁴He → ₁₅³⁰P +₀¹n

₁₅³⁰P → ₁₄³⁰Si + β⁺

In these reactions, ₁₅³⁰P is radioactive nucleus, and it is converted to ₁₄³⁰Si by positron decay. Neutrons ₀¹n, protons ₁¹H, deuterium ₁²H are used in artificial nuclear reactions. Now we explain to important artificial nuclear reactions fission and fusion .

1. Nuclear Fission:

Nuclear fission is a nuclear reaction in which nucleus of atom split into smaller particles. Nucleus having mass number larger than 200 does neutron decay and split into elements having smaller mass numbers.

Example:

₉₂²³⁵U + ₀¹n → ₅₆¹⁴¹Ba + ₃₆⁹²Kr + 3₀¹n + Energy

Nuclear fission is an exothermic reaction and excess amount of energy is released. By the help of these reactions, now energy is produced in nuclear power plants. Picture given below shows fission of uranium;

2. Nuclear Fusion:

More than one nucleus, having small atomic masses, are combined to form heavier new nucleus. Nuclear fusion is also exothermic reactions and energy released in these reactions are larger than energy released in fission reactions. On the contrary, there must be large amount of energy to start fusion reactions.In hydrogen bomb we see fusion reactions.

Example:

₄¹¹H → ₂⁴He + 2β⁺

₁²H + ₁³H → ₂⁴He +₀¹n

Example: Which ones of the following statements are true for nuclear reactions?

I. Sum of mass number is conserved

II. Mass loss is not important

III. Structure of nucleus can change

Solution:

In nuclear reactions, sum of number of protons and neutrons is always conserved. However, in nuclear reactions mass is not conserved. Lost mass is converted to energy, so amount of mass is important. In nuclear reactions one atom can converted to another atom. I and III are true.

Example: Which ones of the following statements are true for following reaction;

₄⁹Be + ₁³H → ₅¹¹B + ₀¹n

I. It is fusion reaction

II. It is natural nuclear reaction

III. Number of neutron is conserved

Solution:

₄⁹Be and ₁³H are joined to form ₅¹¹B. Thus, this is fusion reaction in other words artificial nuclear reaction. If we write number of neutrons in both sides of reaction;

₄⁹Be +₁³H → ₅¹¹B + ₀¹n

5+2=6+1

7=7 Number of neutrons is conserved in this reaction.

Example: Find Z in the reaction given below.

_a³²X + ₂⁴He → _(a+2)³⁵Y + Z

Solution:

_a³²X + ₂⁴He →_(a+2)³⁵Y + _b^cZ

conservation of charge;

a+2=(a+2)+b

b=0

Conservation of mass;

32+4=35+c

c=1

Thus, Z is neutron ₀¹n.

Nuclear Chemistry (Radioactivity) Exams and  Problem Solutions