Chapter 20, Problem 20.66QP

(a)

Interpretation Introduction

Interpretation: For the given species, binding energy per nucleon should be determined.

Concept Introduction:

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is, $\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• The difference between mass of an atom and the sum of the masses of its proton, electron, and neutron is called Mass defect

To determine: Binding energy per nucleon for the given

Answer to Problem 20.66QP

The nuclear binding energy for ${}^{\text{10}}\text{B}$ is $\text{ 1}{\text{.040×10}}^{\text{-12}}\text{J/nucleon}$

Explanation of Solution

For the given ${}^{\text{10}}\text{B}$, there are5 protons and 5 neutrons.

$\begin{array}{l}\text{Mass}\text{of}\text{protons}\text{=}\text{1}{\text{.00728amu}}^{}\\ \text{Mass}\text{of}\text{5}\text{protons}\text{=}\text{1}\text{.00728}\text{×}\text{5=}\text{5}\text{.0364}\text{amu}\\ \text{Mass}\text{of}\text{electrons}\text{=}\text{5}\text{.4858}\text{×}{\text{10}}^{\text{-4}}\text{amu}\\ \text{Mass}\text{of5}\text{electrons}\text{=}\text{5}\text{×}\text{5}\text{.4858}\text{×}{\text{10}}^{\text{-4}}\text{amu=}2.7429\times {10}^{-3}\text{amu}\\ \text{Mass}\text{of}\text{neutron=1}\text{.008665}\text{amu}\\ \text{Mass}\text{of5}\text{neutron}\text{=}5\text{×1}\text{.008665}\text{amu=}5.043325\text{amu}\\ \text{So,}\text{predicted}\text{mass}\text{for}{}^{\text{10}}\text{B}\text{is}\\ \text{5}\text{.0364}\text{amu+}2.7429\times {10}^{-3}\text{amu+}5.043325\text{amu}\\ \text{=}\text{10}\text{.0824679amu}\\ \text{Mass}{\text{of }}^{10}\text{B}\text{is}\text{10}\text{.0129}\text{amu}\\ \text{Mass}\text{defect}\text{=}\text{Atomic}\text{mass}{\text{-M}}_{\text{P}}\text{+}{\text{M}}_{\text{e}}\text{+}{\text{M}}_{\text{n}}\\ \text{=}\text{10}\text{.0129amu}\text{-}10.0824679\text{amu}\\ \text{=}\text{-0}\text{.0695679}\text{amu}\\ \text{Since,}\text{1}\text{kg}\text{=}\text{6}{\text{.022×10}}^{\text{26}}\text{amu}\\ \text{Δm}\text{=}\frac{\text{-0}\text{.0695679}\text{amu}}{\text{6}{\text{.022×10}}^{\text{26}}\text{amu}}\text{=}\text{-1}\text{.15522916×}{\text{10}}^{-28}\text{kg}\end{array}$

The binding energy

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is, $\Delta E=\left(\Delta m\right)c^2$

C=velocity of light is $\text{2}{\text{.99792458×10}}^{\text{8}}\text{m/s}$

$\begin{array}{l}\text{ΔE =}\left(\text{Δm}\right){\text{c}}^{\text{2}}\\ \text{ΔE}\text{=}\text{-1}\text{.15522916 ×}{\text{10}}^{\text{-28}}\text{kg}\text{×}{\left(\text{2}{\text{.99792458×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}\\ \text{=}\text{1}{\text{.03826819×10}}^{\text{-11}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{=}\text{1}{\text{.03826819×10}}^{\text{-11}}\text{J}\\ \text{Nucear}\text{binding}\text{energy}\text{per}\text{nucleon=}\frac{\text{1}{\text{.03826819×10}}^{\text{-11}}J}{\text{10}\text{nucleon}}\\ \text{=}\text{1}\text{.040×}{\text{10}}^{\text{-12}}\text{J/nucleon}\end{array}$

(b)

Interpretation Introduction

Interpretation: For the given species, binding energy per nucleon should be determined.

Concept Introduction:

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is, $\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• The difference between mass of an atom and the sum of the masses of its proton, electron, and neutron is called Mass defect

To determine: Binding energy per nucleon for the given

Answer to Problem 20.66QP

The nuclear binding energy for ${}_{}^{\text{11}}\text{B}$ is $\text{1}\text{.111}{\text{×10}}^{\text{-12}}\text{J/nucleon}$

Explanation of Solution

For the given ${}_{}^{\text{11}}\text{B}$, there are5 protons and 6 neutrons.

$\begin{array}{l}\text{Mass}\text{of}\text{protons}\text{=}\text{1}{\text{.00728amu}}^{}\\ \text{Mass}\text{of}\text{5}\text{protons}\text{=}\text{1}\text{.00728}\text{×}\text{5=}\text{5}\text{.0364}\text{amu}\\ \text{Mass}\text{of}\text{electrons}\text{=}\text{5}\text{.4858}\text{×}{\text{10}}^{\text{-4}}\text{amu}\\ \text{Mass}\text{of5}\text{electrons}\text{=}\text{5}\text{×}\text{5}\text{.4858}\text{×}{\text{10}}^{\text{-4}}\text{amu=}2.7429\times {10}^{-3}\text{amu}\\ \text{Mass}\text{of}\text{neutron=1}\text{.008665}\text{amu}\\ \text{Mass}\text{of6}\text{neutron}\text{=}6\text{×1}\text{.008665}\text{amu=}6.05199\text{amu}\\ \text{So,}\text{predicted}\text{mass}\text{for}{}^{\text{11}}\text{B}\text{is}\\ \text{5}\text{.0364}\text{amu+}2.7429\times {10}^{-3}\text{amu+}6.05199\text{amu}\\ \text{=}\text{11}\text{.0911329amu}\\ \text{Mass}{\text{of }}^{10}\text{B}\text{is}\text{11}\text{.009305}\text{amu}\\ \text{Mass}\text{defect}\text{=}\text{Atomic}\text{mass}{\text{-M}}_{\text{P}}\text{+}{\text{M}}_{\text{e}}\text{+}{\text{M}}_{\text{n}}\\ \text{=}\text{11}\text{.009305amu}\text{-}11.0911329\text{amu}\\ \text{=}\text{-0}\text{.0818279}\text{amu}\\ \text{Since,}\text{1}\text{kg}\text{=}\text{6}{\text{.022×10}}^{\text{26}}\text{amu}\\ \text{Δm}\text{=}\frac{\text{-0}\text{.0818279}\text{amu}}{\text{6}{\text{.022×10}}^{\text{26}}\text{amu}}\text{=}\text{-1}\text{.358816008×}{\text{10}}^{-28}\text{kg}\end{array}$

The binding energy

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is, $\Delta E=\left(\Delta m\right)c^2$

C=velocity of light is $\text{2}{\text{.99792458×10}}^{\text{8}}\text{m/s}$

$\begin{array}{l}\text{ΔE =}\left(\text{Δm}\right){\text{c}}^{\text{2}}\\ \text{ΔE}\text{=}\text{-1}\text{.358816008 ×}{\text{10}}^{\text{-28}}\text{kg}\text{×}{\left(\text{2}{\text{.99792458×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}\\ \text{=}\text{1}{\text{.2212429×10}}^{\text{-11}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{=}\text{1}\text{.2212429}{\text{×10}}^{\text{-11}}\text{J}\\ \text{Nucear}\text{binding}\text{energy}\text{per}\text{nucleon=}\frac{\text{1}\text{.2212429}{\text{×10}}^{\text{-11}}\text{J}}{\text{11}\text{nucleon}}\\ \text{=}\text{1}\text{.11×}{\text{10}}^{\text{-12}}\text{J/nucleon}\end{array}$

(c)

Interpretation Introduction

Interpretation: For the given species, binding energy per nucleon should be determined.

Concept Introduction:

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is, $\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• The difference between mass of an atom and the sum of the masses of its proton, electron, and neutron is called Mass defect

To determine: Binding energy per nucleon for the given

Answer to Problem 20.66QP

The nuclear binding energy for ${}^{\text{14}}\text{N }$ is $\text{1}{\text{.199×10}}^{\text{-12}}\text{J/nucleon}$

Explanation of Solution

For the given ${}^{\text{14}}\text{N}$, there are7 protons and 7 neutrons.

$\begin{array}{l}\text{Mass}\text{of}\text{protons}\text{=}\text{1}{\text{.00728amu}}^{}\\ \text{Mass}\text{of}7\text{protons}\text{=}\text{1}\text{.00728}\text{×}7\text{=}7.05096\text{amu}\\ \text{Mass}\text{of}\text{electrons}\text{=}\text{5}\text{.4858}\text{×}{\text{10}}^{\text{-4}}\text{amu}\\ \text{Mass}\text{of7}\text{electrons}\text{=}\text{7×}\text{5}\text{.4858}\text{×}{\text{10}}^{\text{-4}}\text{amu=}3.84006\times {10}^{-3}\text{amu}\\ \text{Mass}\text{of}\text{neutron=1}\text{.008665}\text{amu}\\ \text{Mass}\text{of7}\text{neutron}\text{=}7\text{×1}\text{.008665}\text{amu=}7.060655\text{amu}\\ \text{So,}\text{predicted}\text{mass}\text{for}{}^{\text{14}}N\text{is}\\ 7.05096\text{amu+}3.84006\times {10}^{-3}\text{amu+}7.060655\text{amu}\\ \text{=}14.11545506\text{amu}\\ \text{Mass}{\text{of }}^{14}\text{N}\text{is}\text{14}\text{.003074}\text{amu}\\ \text{Mass}\text{defect}\text{=}\text{Atomic}\text{mass}{\text{-M}}_{\text{P}}\text{+}{\text{M}}_{\text{e}}\text{+}{\text{M}}_{\text{n}}\\ \text{=}\text{14}\text{.003074amu}\text{-}14.11545506\text{amu}\\ \text{=}\text{-0}\text{.11238106amu}\\ \text{Since,}\text{1}\text{kg}\text{=}\text{6}{\text{.022×10}}^{\text{26}}\text{amu}\\ \text{Δm}\text{=}\frac{\text{-0}\text{.11238106amu}}{\text{6}{\text{.022×10}}^{\text{26}}\text{amu}}\text{=}\text{-1}\text{.866175025×}{\text{10}}^{-28}\text{kg}\end{array}$

The binding energy

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is, $\Delta E=\left(\Delta m\right)c^2$

C=velocity of light is $\text{2}{\text{.99792458×10}}^{\text{8}}\text{m/s}$

$\begin{array}{l}\text{ΔE =}\left(\text{Δm}\right){\text{c}}^{\text{2}}\\ \text{ΔE}\text{=}\text{-1}\text{.866175 ×}{\text{10}}^{\text{-28}}\text{kg}\text{×}{\left(\text{2}{\text{.99792458×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}\\ \text{=}\text{1}{\text{.677234468×10}}^{\text{-11}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{=}\text{1}\text{.677234468}{\text{×10}}^{\text{-11}}\text{J}\\ \text{Nucear}\text{binding}\text{energy}\text{per}\text{nucleon=}\frac{\text{1}{\text{.677234468×10}}^{\text{-11}}\text{J}}{\text{14}\text{nucleon}}\\ \text{=}\text{1}\text{.199×}{\text{10}}^{\text{-12}}\text{J/nucleon}\end{array}$

(d)

Interpretation Introduction

Interpretation: For the given species, binding energy per nucleon should be determined.

Concept Introduction:

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is, $\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• The difference between mass of an atom and the sum of the masses of its proton, electron, and neutron is called Mass defect

To determine: Binding energy per nucleon for the given

Answer to Problem 20.66QP

The nuclear binding energy for ${}^{\text{56}}\text{Fe}$ is $\text{1}{\text{.410×10}}^{\text{-12}}\text{J/nucleon}$

Explanation of Solution

For the given ${}^{\text{56}}\text{Fe}$, there are26 protons and 30 neutrons.

$\begin{array}{l}\text{Mass}\text{of}\text{protons}\text{=}\text{1}{\text{.00728amu}}^{}\\ \text{Mass}\text{of}\text{26protons}\text{=}\text{1}\text{.00728}\text{×}\text{26=}\text{26}\text{.18928}\text{amu}\\ \text{Mass}\text{of}\text{electrons}\text{=}\text{5}\text{.4858}\text{×}{\text{10}}^{\text{-4}}\text{amu}\\ \text{Mass}\text{of26}\text{electrons}\text{=}\text{26×}\text{5}\text{.4858}\text{×}{\text{10}}^{\text{-4}}\text{amu=}\text{0}\text{.01426308amu}\\ \text{Mass}\text{of}\text{neutron=1}\text{.008665}\text{amu}\\ \text{Mass}\text{of30}\text{neutron}\text{=}\text{30×1}\text{.008665}\text{amu=}\text{30}\text{.25995amu}\\ \text{So,}\text{predicted}\text{mass}\text{for}{}^{\text{56}}\text{Fe}\text{is}\\ \text{26}\text{.18928}\text{amu+0}\text{.01426308amu}\text{+}\text{30}\text{.25995amu}\\ \text{=}\text{56}\text{.4634amu}\\ \text{Mass}{\text{of }}^{\text{56}}\text{Fe}\text{is}\text{55}\text{.93494}\text{amu}\\ \text{Mass}\text{defect}\text{=}\text{Atomic}\text{mass}{\text{-M}}_{\text{P}}\text{+}{\text{M}}_{\text{e}}\text{+}{\text{M}}_{\text{n}}\\ \text{=}\text{55}\text{.93494amu}\text{-}\text{56}\text{.4634amu}\\ \text{=}\text{-0}\text{.52855308amu}\\ \text{Since,}\text{1}\text{kg}\text{=}\text{6}{\text{.022×10}}^{\text{26}}\text{amu}\\ \text{Δm}\text{=}\frac{\text{-0}\text{.52855308amu}}{\text{6}{\text{.022×10}}^{\text{26}}\text{amu}}\text{=}\text{-8}\text{.77698×}{\text{10}}^{\text{-28}}\text{kg}\end{array}$

The binding energy

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is, $\Delta E=\left(\Delta m\right)c^2$

C=velocity of light is $\text{2}{\text{.99792458×10}}^{\text{8}}\text{m/s}$

$\begin{array}{l}\text{ΔE =}\left(\text{Δm}\right){\text{c}}^{\text{2}}\\ \text{ΔE}\text{=}\text{-8}\text{.77698×}{\text{10}}^{\text{-28}}\text{kg}\text{×}{\left(\text{2}{\text{.99792458×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}\\ \text{=}-7.887475{\text{×10}}^{\text{-11}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{=}-7.887475{\text{×10}}^{\text{-11}}\text{J}\\ \text{Nucear}\text{binding}\text{energy}\text{per}\text{nucleon=}\frac{-7.887475{\text{×10}}^{\text{-11}}\text{J}}{56\text{nucleon}}\\ \text{=}-1.40\text{×}{\text{10}}^{\text{-12}}\text{J/nucleon}\end{array}$