Chapter 1, Problem 1.95SP

(a)

Interpretation Introduction

Interpretation:

The particle produced in the given decay reaction has to be identified.

Concept introduction:

• Beta decay is the emission of an electron and the simultaneous conversion of a neutron into a proton in the nucleus.  So increase the atomic number by one.
• When the beta emission a neutron is converted to a proton and the atomic number increases.

Beta Decay:   ${}_{\text{6}}^{\text{14}}\text{C}\to \text{N}\text{+}\text{e}$

(b)

Interpretation Introduction

Interpretation:

The particle produced in the given decay reaction has to be identified.

Concept introduction:

Positron Decay:

• Positron emission is outgoing particle has positive charge and a proton is converted into a neutron.  It decreases the atomic number.
• Positron Decay:  ${}_{\text{12}}^{\text{23}}\text{Mg}\to \text{N}\text{+}\text{e}$
• In positron emission a proton is converted to a neutron and the atomic number decreases.

(c)

Interpretation Introduction

Interpretation:

The particle produced in the given decay reaction has to be identified.

Concept introduction:

• The nuclear disintegration process that emits alpha particles is called alpha decay. An example of a nucleus that undergoes alpha decay is ${}^{\text{238}}\text{U}$.  The alpha decay of ${}^{\text{238}}\text{U}$
• Alpha Decay:   ${}_{\text{92}}^{\text{238}}\text{U}\to \text{H}\text{e}\text{+}\text{T}\text{h}$