Chapter 25.1, Problem 6LC

Interpretation Introduction

Interpretation: To compare the effect of mass number during an alpha and beta decay.

Concept Introduction: The process which involves the formation of more than one nuclide via a collision between two nuclei or between one nucleus and a subatomic particle, is known as a nuclear reaction.

Answer to Problem 6LC

In alpha decay, the mass number is decreased by four units while in beta decay, there is no change in the mass number.

Explanation of Solution

Alpha decay: The radiation that emits an alpha particle (Helium nuclei) is known as alpha radiation. An alpha particle comprises two protons and two neutrons and also consists of a double positive layer.

The chemical equation depicting alpha radiation is as follows:

  $\text{U}\stackrel{\text{radioactive decay}}{\to }\text{T}\text{h}+\text{H}\text{e}$

From the above chemical equation, it is seen that in alpha decay, the atomic number (number written in subscripts) of the product decreases by two while the mass number (number written in superscript) is decreased by four units.

Beta decay: When an electron is formed that results from breaking a neutron, this electron is termed a beta particle. The neutron breaks into a proton and an electron is released in the beta radiation.

The chemical equation depicting beta radiation is as follows:

  $\text{C}\to \text{N}+\text{e}$

In beta decay, the atomic number of the product decreases by one while the mass number remains the same.

Conclusion

In alpha decay, the mass number is decreased by four units while in beta decay, there is no change in the mass number.