Chapter 10, Problem 10C.7E

(a)

Interpretation Introduction

Interpretation:

The nuclear equation in which sodium-11 decays into magnesium-12 have to be given.

Concept Introduction:

Beta decay reaction:

Beta particles are electrons (electrons with positive electric charge).  Beta decay occurs when in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other.

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

Answer to Problem 10C.7E

The nuclear equation is,

  ${}_{\text{11}}{\text{Na}}^{\text{24}}{\stackrel{}{\to }}_{\text{12}}{\text{Mg}}^{\text{24}}{\text{+}}_{\text{-1}}{\text{e}}^{\text{0}}$

Explanation of Solution

Sodium-11 decays into magnesium-12 by emitting an electron particle.

The nuclear equation is,

  ${}_{\text{11}}{\text{Na}}^{\text{24}}{\stackrel{}{\to }}_{\text{12}}{\text{Mg}}^{\text{24}}{\text{+}}_{\text{-1}}{\text{e}}^{\text{0}}$

(b)

Interpretation Introduction

Interpretation:

The change in energy that accompanies the decay has to be given.

Concept Introduction:

Nuclear binding energy and the mass defect:

Nuclear binding energy is the minimum energy that would be required to disassemble the nucleus of an atom into its component parts.  These component parts are neutrons and protons which are collectively called nucleons.  The mass of an atomic nucleus is less than the sum of the individual masses of the free constituent protons and neutrons according to Einstein’s equation ${\text{E=mc}}^{\text{2}}$.  The missing mass is known as the mass defect and represents the energy that was released when the nucleus was formed.

  $\Delta \text{m=}\sum \text{m}\left(\text{products}\right)\text{-}\sum \text{m}\left(\text{reactants}\right)$

The rate of energy is calculated using the formula,

  $\text{ΔE=}{\text{-Δm×c}}^{\text{2}}$

Where $\text{ΔE}$ is the rate of energy.

$\text{Δm}$ is the rate of mass loss.

c is the velocity of the light $\left({\text{3×10}}^{\text{8}}\text{m}{\text{s}}^{\text{-1}}\right)$

Answer to Problem 10C.7E

The change in energy that accompanies the decay is $\text{-0}{\text{.08847×10}}^{\text{-11}}\text{J}\text{.}$

Explanation of Solution

Given,

The mass of sodium is $\text{23}\text{.99096mu}\text{.}$

The mass of magnesium is $\text{23}\text{.98504mu}\text{.}$

The mass of electron is $\text{0}\text{.00054858mu}\text{.}$

The change in mass can be calculated as,

  $\Delta \text{m=}\sum \text{m}\left(\text{products}\right)\text{-}\sum \text{m}\left(\text{reactants}\right)$

  $\Delta \text{m=}\sum \text{m}\left(23.98504\right)\text{-}\sum \text{m}\left(23.99096\right)$

  $\text{Δm}\text{=}\text{-0}\text{.00592mu}\text{.}$

The change in energy can be calculated as,

  $\text{ΔE=}{\text{-Δm×c}}^{\text{2}}$

  $\text{ΔE=}\text{-0}\text{.00592mu}\text{.}\left({\text{9×10}}^{\text{16}}\text{m}{\text{s}}^{\text{-1}}\right)$

  $\text{ΔE}\text{=-}\text{0}{\text{.05328×10}}^{\text{16}}\text{m}{\text{s}}^{\text{-1}}$

  $\text{ΔE}\text{=-}\text{0}{\text{.05328×10}}^{\text{16}}\text{m}{\text{s}}^{\text{-1}}\text{×1}{\text{.66054×10}}^{\text{-27}}\text{kg}$

  $\text{ΔE}\text{=}\text{-0}{\text{.08847×10}}^{\text{-11}}\text{J}$

The change in energy is $\text{-0}{\text{.08847×10}}^{\text{-11}}\text{J}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The change in energy per nucleon has to be given.

Concept Introduction:

The binding energy per nucleon (BEN) is defined by the average energy required to remove an individual nucleus from the nucleus.

  $\text{BEN=}\frac{\text{ΔE}}{\text{A}}$

Where $\text{ΔE}$ is the change in energy.

A is the mass number of the atom.

Answer to Problem 10C.7E

The binding energy per nucleon is $\text{0}{\text{.00368×10}}^{\text{-11}}\text{J /nucleon}$.

Explanation of Solution

Given,

The change in energy is $\text{-0}{\text{.08847×10}}^{\text{-11}}\text{J}\text{.}$

The mass umber of the sodium atom is $\text{23}\text{.99096mu}\text{.}$

The binding energy per nucleon can be calculated as,

  $\text{BEN=}\frac{\text{ΔE}}{\text{A}}$

  $\text{BEN=}\frac{\text{0}{\text{.08847×10}}^{\text{-11}}\text{J}}{\text{23}\text{.99096}}$

  $\text{BEN=0}{\text{.00368×10}}^{\text{-11}}\text{J /nucleon}$

The binding energy per nucleon is $\text{0}{\text{.00368×10}}^{\text{-11}}\text{J /nucleon}$.