Chapter 10, Problem 10C.6E

(a)

Interpretation Introduction

Interpretation:

The amount of energy released 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.

  $\Delta \text{m=mass of the products -mass of the reactants}$

Answer to Problem 10C.6E

The total amount energy released is $\text{E=128}\text{.4}\times {\text{10}}^{12}\text{J}\text{.}$

Explanation of Solution

Given:

The mass of neutron is $\text{1}\text{.008664u}\text{.}$

The mass of Beryllium is $\text{7}\text{.0169mu}\text{.}$

The mass of the hydrogen-1 is $\text{0}\text{.0078mu}\text{.}$

The mass of Lithium is $\text{7}\text{.0160mu}\text{.}$

The mass defect can be calculated as,

  $\Delta \text{m=mass of the products -mass of the reactants}$

  $\begin{array}{l}\Delta \text{m=}1.0017\text{×1}{\text{.66054×10}}^{\text{-27}}\\ \text{=}1.6634{\text{×10}}^{\text{-27}}\text{kg}\end{array}$

The change in energy ($\Delta \text{E}$) =$\Delta {\text{m×c}}^{\text{2}}$

  ($\Delta \text{E}$)=$\text{1}{\text{.6634×10}}^{\text{-27}}\text{×}{\left({\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}$

  ($\Delta \text{E}$)=$\text{14}{\text{.97×10}}^{\text{-11}}\text{J}\text{.}$

The number of atoms in the sample is,

  $\begin{array}{l}\text{N=}\frac{\text{mass}\text{of}\text{the}\text{sample}}{\text{mass}\text{of}\text{the}\text{atom}}\\ \text{N=}\frac{\text{1}{\text{.00×10}}^{\text{-3}}\text{kg}}{\text{7}\text{.0160×1}{\text{.66054×10}}^{\text{-27}}\text{kg}}\\ \text{N=8}\text{.58}\times {\text{10}}^{23}\end{array}$

The total amount of energy released is,

  $\begin{array}{l}\text{E=NΔE}\\ \text{E=8}{\text{.58×10}}^{\text{23}}\left(\text{14}{\text{.97×10}}^{\text{-11}}\text{J}\right)\\ \text{E=128}{\text{.4×10}}^{\text{12}}\text{J}\end{array}$

The total amount energy released is $\text{E=128}\text{.4}\times {\text{10}}^{12}\text{J}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The amount of energy released has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10C.6E

The total amount energy released is $\text{E=1}\text{.5065}\times {\text{10}}^{12}\text{J}\text{.}$

Explanation of Solution

Given:

The mass of Hydrogen-1 is $\text{0}\text{.0078u}\text{.}$

The mass of Cobalt-60 is $\text{59}\text{.959mu}\text{.}$

The mass of the Cobalt-59 is $\text{58}\text{.93332mu}\text{.}$

The mass of Deuterium is $\text{2}\text{.0414mu}\text{.}$

The mass defect can be calculated as,

  $\Delta \text{m=mass of the products -mass of the reactants}$

  $\begin{array}{l}\Delta \text{m=}\text{-0}\text{.9866×1}{\text{.66054×10}}^{\text{-27}}\\ \text{=}\text{-}\text{1}{\text{.68382×10}}^{\text{-27}}\text{kg}\end{array}$

The change in energy ($\Delta \text{E}$) =$\Delta {\text{m×c}}^{\text{2}}$

  ($\Delta \text{E}$)=$\text{1}{\text{.6838×10}}^{\text{-27}}\text{×}{\left({\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}$

  ($\Delta \text{E}$)=$\text{1}{\text{.4744×10}}^{\text{-10}}\text{J}\text{.}$

The number of atoms in the sample is,

  $\begin{array}{l}\text{N=}\frac{\text{mass}\text{of}\text{the}\text{sample}}{\text{mass}\text{of}\text{the}\text{atom}}\\ \text{N=}\frac{\text{1}{\text{.00×10}}^{\text{-3}}\text{kg}}{\text{58}\text{.9332×1}{\text{.66054×10}}^{\text{-27}}\text{kg}}\\ \text{N=1}\text{.02185}\times {\text{10}}^{23}\end{array}$

The total amount of energy released is,

  $\begin{array}{l}\text{E=NΔE}\\ \text{E=1}{\text{.02185×10}}^{\text{23}}\left(\text{1}{\text{.4744×10}}^{\text{-10}}\text{J}\right)\\ \text{E=1}{\text{.5065×10}}^{\text{12}}\text{J}\end{array}$

The total amount energy released is $\text{E=1}\text{.5065}\times {\text{10}}^{12}\text{J}.$

(c)

Interpretation Introduction

Interpretation:

The amount of energy released has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10C.6E

The total amount energy released is $\text{E=0}\text{.011}\times {\text{10}}^{12}\text{J}\text{.}$

Explanation of Solution

Given:

The mass of Krypton is $\text{39}\text{.9640mu}\text{.}$

The mass of Argon is $\text{39}\text{.9640mu}\text{.}$

The mass of Beta particle is $\text{0}\text{.0005mu}\text{.}$

The mass defect can be calculated as,

  $\Delta \text{m=mass of the products -mass of the reactants}$

  $\begin{array}{l}\Delta \text{m=}\text{-0}\text{.0005×1}{\text{.66054×10}}^{\text{-27}}\\ \text{=}0.00083{\text{×10}}^{\text{-27}}\text{kg}\end{array}$

The change in energy ($\Delta \text{E}$) =$\Delta {\text{m×c}}^{\text{2}}$

  ($\Delta \text{E}$)=$\text{0}{\text{.000083×10}}^{\text{-27}}\text{×}{\left({\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}$

  ($\Delta \text{E}$)=$\text{0}{\text{.0074×10}}^{\text{-11}}\text{J}\text{.}$

The number of atoms in the sample is,

  $\begin{array}{l}\text{N=}\frac{\text{mass}\text{of}\text{the}\text{sample}}{\text{mass}\text{of}\text{the}\text{atom}}\\ \text{N=}\frac{\text{1}{\text{.00×10}}^{\text{-3}}\text{kg}}{\text{39}\text{.9640×1}{\text{.66054×10}}^{\text{-27}}\text{kg}}\\ \text{N=1}\text{.50}\times {\text{10}}^{23}\end{array}$

The total amount of energy released is,

  $\begin{array}{l}\text{E=NΔE}\\ \text{E=1}{\text{.50×10}}^{\text{23}}\left(\text{0}{\text{.0074×10}}^{\text{-11}}\text{J}\right)\\ \text{E=0}{\text{.011×10}}^{\text{12}}\text{J}\text{.}\end{array}$

The total amount energy released is $\text{E=0}\text{.011}\times {\text{10}}^{12}\text{J}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The total amount of energy released has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10C.6E

The total amount energy released is $\text{E=1}\text{.5312}\times {\text{10}}^{12}\text{J}\text{.}$

Explanation of Solution

Given:

The mass of Boron is $\text{10}\text{.0129mu}\text{.}$

The mass of neutron is $\text{1}\text{.008664u}\text{.}$

The mass of Helium-4 is $\text{4}\text{.0026mu}\text{.}$

The mass of Lithium is $\text{7}\text{.0160mu}\text{.}$

The mass defect can be calculated as,

  $\Delta \text{m=mass of the products -mass of the reactants}$

  $\begin{array}{l}\Delta \text{m=}\text{-0}\text{.0170×1}{\text{.66054×10}}^{\text{-27}}\\ \text{=}0.0282{\text{×10}}^{\text{-27}}\text{kg}\end{array}$

The change in energy ($\Delta \text{E}$) =$\Delta {\text{m×c}}^{\text{2}}$

  ($\Delta \text{E}$)=$\text{0}{\text{.0282×10}}^{\text{-27}}\text{×}{\left({\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}$

  ($\Delta \text{E}$)=$\text{0}{\text{.2546×10}}^{\text{-11}}\text{J}\text{.}$

The number of atoms in the sample is,

  $\begin{array}{l}\text{N=}\frac{\text{mass}\text{of}\text{the}\text{sample}}{\text{mass}\text{of}\text{the}\text{atom}}\\ \text{N=}\frac{\text{1}{\text{.00×10}}^{\text{-3}}\text{kg}}{\text{2}\text{.0141×1}{\text{.66054×10}}^{\text{-27}}\text{kg}}\\ \text{N=6}\text{.0145}\times {\text{10}}^{22}\end{array}$

The total amount of energy released is,

  $\begin{array}{l}\text{E=NΔE}\\ \text{E=6}{\text{.0145×10}}^{\text{22}}\left(\text{0}{\text{.2546×10}}^{\text{-11}}\text{J}\right)\\ \text{E=1}{\text{.5312×10}}^{\text{12}}\text{J}\text{.}\end{array}$

The total amount energy released is $\text{E=1}\text{.5312}\times {\text{10}}^{12}\text{J}\text{.}$