Chapter 24, Problem 24.80P

(a)

Interpretation Introduction

Interpretation:

Binding energy per nucleon of ${}^{206}Pb$ has to be calculated in $MeV.$

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  $\text{Nucleus + nuclear binding energy }\underset{}{\to }\text{ nucleons}$

It can be calculated in $MeV$ using the given formula,

  $\text{Binding energy = }\Delta \text{m}\left(\text{in amu}\right)\times \frac{\text{931}\text{.5 MeV}}{1\text{ amu}}$

Nucleon: Nucleon is the collective name given to the members of the nucleus that is protons and neutrons.

The difference between mass of an atom and the sum of the masses of its proton and neutron is called Mass difference.

Binding Energy per Nucleon can be determined as follows:

  $\text{Binding energy per nucleon = }\frac{\text{Binding energy}}{\text{No}\text{. of nucleons}}$

Explanation of Solution

Given information is shown below,

Mass of Lead-206 is $\text{205}\text{.974440 amu}.$

In ${}^{206}Pb$, there are 82 protons and 124 neutrons.

• Calculate the mass difference:

As known the mass $H$ atom is $\text{1}\text{.007825 amu}$ and mass of neutron is $\text{1}\text{.008665 amu}\text{.}$

Mass difference in ${}^{206}Pb$ can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }\left[\left(82\times \text{mass H atom}\right)+\left(124\times \text{mass of neutron}\right)\right]-\left[\text{205}\text{.974440 amu}\right]\\ \\ =\left[\left(82\times \text{1}\text{.007825 amu}\right)+\left(124\times \text{1}\text{.008665 amu}\right)\right]-\left[\text{205}\text{.974440 amu}\right]\\ \\ =\left[\left(\text{82}\text{.641650 amu}\right)+\left(\text{125}\text{.074460 amu}\right)\right]-\left[\text{205}\text{.974440 amu}\right]\\ \\ =\text{ 1}\text{.741670 amu}\end{array}$

Mass difference in ${}^{206}Pb$ is $\text{1}\text{.741670 amu}\text{.}$

• Calculate the binding energy per nucleon:

Number of nucleons in ${}^{206}Pb$ is 206.

Binding energy per nucleon of ${}^{206}Pb$ in $MeV$ is calculated as follows,

  $\begin{array}{c}BindingEnergyperNucleon=\frac{\Delta \text{m}\left(\text{in amu}\right)\times \frac{\text{931}\text{.5 MeV}}{1\text{ amu}}}{\text{No}\text{. of nucleons}}\\ \\ =\frac{\left(\text{1}\text{.741670 amu}\right)\times \frac{\text{931}\text{.5 MeV}}{1\text{ amu}}}{\text{206 nucleons}}\\ \\ =\text{ 7}\text{.876 MeV/nucleon}\end{array}$

Binding energy per nucleon of ${}^{206}Pb$ is $7.876MeV/nucleon.$

(b)

Interpretation Introduction

Interpretation:

Binding energy per atom of ${}^{206}Pb$ has to be calculated in $MeV.$

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  $\text{Nucleus + nuclear binding energy }\underset{}{\to }\text{ nucleons}$

It can be calculated in $MeV$ using the given formula,

  $\text{Binding energy = }\Delta \text{m}\left(\text{in amu}\right)\times \frac{\text{931}\text{.5 MeV}}{1\text{ amu}}$

The difference between mass of an atom and the sum of the masses of its proton and neutron is called Mass difference.

Explanation of Solution

Given information is shown below,

Mass of Lead-206 is $\text{205}\text{.974440 amu}.$

In ${}^{206}Pb$, there are 82 protons and 124 neutrons.

• Calculate the mass difference:

As known the mass $H$ atom is $\text{1}\text{.007825 amu}$ and mass of neutron is $\text{1}\text{.008665 amu}\text{.}$

Mass difference in ${}^{206}Pb$ can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }\left[\left(82\times \text{mass H atom}\right)+\left(124\times \text{mass of neutron}\right)\right]-\left[\text{205}\text{.974440 amu}\right]\\ \\ =\left[\left(82\times \text{1}\text{.007825 amu}\right)+\left(124\times \text{1}\text{.008665 amu}\right)\right]-\left[\text{205}\text{.974440 amu}\right]\\ \\ =\left[\left(\text{82}\text{.641650 amu}\right)+\left(\text{125}\text{.074460 amu}\right)\right]-\left[\text{205}\text{.974440 amu}\right]\\ \\ =\text{ 1}\text{.741670 amu}\end{array}$

Mass difference in ${}^{206}Pb$ is $\text{1}\text{.741670 amu}\text{.}$

• Calculate the binding energy per atom:

Binding energy per atom of ${}^{206}Pb$ in $MeV$ is calculated as follows,

  $\begin{array}{c}BindingEnergyperAtom=\frac{\Delta \text{m}\left(\text{in amu}\right)\times \frac{\text{931}\text{.5 MeV}}{1\text{ amu}}}{\text{Atom}}\\ \\ =\frac{\left(\text{1}\text{.741670 amu}\right)\times \frac{\text{931}\text{.5 MeV}}{1\text{ amu}}}{1\text{ atom}}\\ \\ =\text{ 1622 MeV/atom}\end{array}$

Binding energy per atom of ${}^{206}Pb$ is $1622MeV/atom.$

(c)

Interpretation Introduction

Interpretation:

Binding energy per mole of ${}^{206}Pb$ has to be calculated in $kJ.$

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  $\text{Nucleus + nuclear binding energy }\underset{}{\to }\text{ nucleons}$

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,

  $\begin{array}{l}\Delta \text{E = }\Delta {\text{mc}}^2\\ \\ where,\\ \Delta \text{m = Mass Difference}\\ c=\text{ Speed of light}\end{array}$

The difference between mass of an atom and the sum of the masses of its proton and neutron is called Mass difference.

Explanation of Solution

Given information is shown below,

Mass of Lead-206 is $\text{205}\text{.974440 amu}.$

In ${}^{206}Pb$, there are 82 protons and 124 neutrons.

• Calculate the mass difference:

As known the mass $H$ atom is $\text{1}\text{.007825 amu}$ and mass of neutron is $\text{1}\text{.008665 amu}\text{.}$

Mass difference in ${}^{206}Pb$ can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }\left[\left(82\times \text{mass H atom}\right)+\left(124\times \text{mass of neutron}\right)\right]-\left[\text{205}\text{.974440 amu}\right]\\ \\ =\left[\left(82\times \text{1}\text{.007825 amu}\right)+\left(124\times \text{1}\text{.008665 amu}\right)\right]-\left[\text{205}\text{.974440 amu}\right]\\ \\ =\left[\left(\text{82}\text{.641650 amu}\right)+\left(\text{125}\text{.074460 amu}\right)\right]-\left[\text{205}\text{.974440 amu}\right]\\ \\ =\text{ 1}\text{.741670 amu = 1}\text{.741670 g/mol}\end{array}$

Mass difference in ${}^{206}Pb$ is $\text{1}\text{.741670 g/mol}\text{.}$

• Calculate the binding energy per mole:

Binding energy per mole of ${}^{206}Pb$ in $kJ$ is calculated as follows,

$\begin{array}{c}\text{Binding energy, E = }\Delta {\text{mc}}^2\\ \\ =\left(\frac{\text{1}\text{.741670 g}}{1\text{ mol}}\right)\left(\frac{1\text{ kg}}{{10}^3\text{ g}}\right){\left(2.998\times {10}^8\text{ m/s}\right)}^2\left(\frac{1\text{ J}}{1\text{ kg}{\text{.m}}^2.s^{-2}}\right)\left(\frac{1\text{ kg}}{{10}^3\text{ J}}\right)\\ \\ =\text{ 1}\text{.56533}\times {\text{10}}^{11}\text{ kJ/mol}\end{array}$

Binding energy per mole of ${}^{206}Pb$ is $1.56533\times 10^{11}kJ/mol.$