Chapter 10, Problem 10A.1AST

(a)

Interpretation Introduction

Interpretation:

The nuclide produced when electron captured by beryllium-7 has to be identified and the nuclear reaction has to be given.

Concept Introduction:

Electron capture is one process that unstable atoms can use to become more stable.  During electron capture, an electron in an atoms inner shell is drawn into the nucleus where it combines with a proton forming a neutron and a neutrino.  The neutrino is ejected from the atoms nucleus

  ${}_{\text{Z}}{\text{X}}^{\text{A}}{\text{+}}_{\text{-1}}{\text{e}}^{\text{0}}{\stackrel{}{\to }}_{\text{Z-1}}{\text{X}}^{\text{A}}$

Answer to Problem 10A.1AST

The nuclide produced is identified as lithium-7and the nuclear reaction is,

  ${}_{\text{4}}{\text{Be}}^{\text{7}}{\text{+}}_{\text{-1}}{\text{e}}^{\text{0}}{\stackrel{}{\to }}_{\text{3}}{\text{Li}}^{\text{7}}$

Explanation of Solution

The nuclear reaction is,

  ${}_{\text{4}}{\text{Be}}^{\text{7}}{\text{+}}_{\text{-1}}{\text{e}}^{\text{0}}{\stackrel{}{\to }}_{\text{3}}{\text{Li}}^{\text{7}}$

In electron capture reaction, a proton has been converted into a neutron, so there was a decrease in atomic number but no change in mass.

(b)

Interpretation Introduction

Interpretation:

The nuclide produced when positron emission by sodium -22 has to be identified and the nuclear reaction has to be given.

Concept Introduction:

Positron emission or ${\text{β}}^{\text{+}}$ decay is a type of radioactive decay called beta decay, in which a proton inside a radioactive nucleus is converted into neutron while releasing a positron and an electron neutrino.  Positron emission is mediated by the weak force.

  ${}_{\text{Z}}{\text{X}}^{\text{A}}\text{+}{\stackrel{}{\to }}_{\text{Z-1}}{\text{X}}^{\text{A}}{\text{+}}_{\text{+1}}{\text{e}}^{\text{0}}$

Answer to Problem 10A.1AST

The nuclide produced is identified as Neon-10 the nuclear reaction is,

  ${}_{\text{11}}{\text{Na}}^{\text{22}}\text{+}{\stackrel{}{\to }}_{\text{10}}{\text{Ne}}^{\text{22}}{\text{+}}_{\text{+1}}{\text{e}}^{\text{0}}$

Explanation of Solution

The nuclear reaction is,

  ${}_{\text{11}}{\text{Na}}^{\text{22}}\text{+}{\stackrel{}{\to }}_{\text{10}}{\text{Ne}}^{\text{22}}{\text{+}}_{\text{+1}}{\text{e}}^{\text{0}}$

In positron emission reaction, a proton has been converted into a neutron, so there was a decrease in atomic number but no change in mass