Chapter 20, Problem 19QP

Interpretation Introduction

Interpretation:

The nuclear binding energy and nuclear binding energyper nucleon, in joules, for a given nucleus is to be determined.

Concept introduction:

The nuclear binding energy is calculated by using the relation as follows:

$\Delta E=\left(\Delta m\right){\text{c}}^2$

Here, $\Delta m$ is the mass defect and c is the speed of light.

The mass defect is defined as the difference between the given mass and the predicted mass. Thus, the formula of themass defect is as follows:

$\Delta m=\text{ Given mass }-\text{Predicted mass}$

Here, $\Delta m$ is the mass defect.

Answer to Problem 19QP

Solution:

$6.30\times {10}^{-12}\text{ J }$, $9.00\times {10}^{-13}\text{ J/nucleon}$

$4.78\times {10}^{-11}\text{ J }$, $1.37\times {10}^{-12}\text{ J/nucleon}.$

Explanation of Solution

a) ${}_3^7\text{Li}$ $\text{7}\text{.01600 amu}$.

In the given nucleus, ${}_3^7\text{Li}$, the number of protons and neutrons are $3\text{ and }4$, respectively.

The mass of $3$ ${}_1^1\text{H}$ atoms is as follows:

$\begin{array}{c}{\text{Mass of 3 }}_1^1\text{H }=3\times \text{mass of hydrogen}\\ =\left(3\times 1.007825\text{ amu}\right)\\ =3.023475\text{ amu}\end{array}$

$\begin{array}{c}\text{Mass of }4\text{ neutrons }=\left(4\times \text{mass of neutron}\right)\\ =\left(\text{4×1}\text{.008665 amu}\right)\\ =4.\text{03466 amu}\end{array}$

The predicted mass of ${}_3^7\text{Li}$ is calculated as follows:

$\begin{array}{c}\text{Predicted mass = Mass of }3{}_1^1\text{H}+\text{mass of }4\text{ neutrons}\\ =\left(3.023475\text{ amu}+4.03466\text{ amu}\right)\\ =7.058315\text{ amu}\end{array}$

The mass defect is calculated by using the relation as follows:

$\Delta m=\text{Given mass }-\text{Predicted mass}$

Substitute $7.058315\text{ amu}$ for predicted mass and $\text{7}\text{.01600 amu}$ for a given mass.

$\begin{array}{c}\Delta \text{m}=\left(\text{7}\text{.01600 amu}\right)-\left(7.058315\text{ amu}\right)\\ =-0.042135\text{ amu}\text{.}\end{array}$

The value of mass defect is converted into Kg by using the relation:

$\begin{array}{c}1\text{ Kg}=6.0221418\times {10}^{26}\text{ amu}\\ \left(-0.042135\cancel{\text{amu}}\times \frac{1\text{ Kg}}{6.022\times {10}^{26}\cancel{\text{amu}}}\right)=-6.996\times {10}^{-29}\text{Kg}\end{array}$

The nuclear binding energy is calculated by the relation as follows:

$\Delta \text{E=}\left(\Delta \text{m}\right){\text{c}}^2$

Substitute $-6.996\times {10}^{-29}\text{Kg}$ for $\Delta \text{m}$ and $3.00\times {10}^8\text{ m/s}$ for c in the above expression.

$\begin{array}{c}\Delta \text{E=}\left(-6.996\times {10}^{-29}\text{Kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2\\ =-63.0\times {10}^{-13}\text{ Kg}{\text{.m}}^2/{\text{s}}^2\\ =-6.30\times {10}^{-12}\text{ J }\end{array}$

The magnitude of binding energy is considered.

Now, the nuclear binding energy per nucleon is calculated as follows:

$\frac{\text{Nuclear binding energy}}{\text{total number of nucleons}}$

The number of nucleons is defined as the sum of protons and neutrons.

Thus, $7$ nucleons are present in nucleus ${}_3^7\text{Li}$.

Substitute $6.30\times {10}^{-12}\text{ J }$ for nuclear binding energy and $7$ nucleons for the total number of nucleons,

$\frac{6.30\times {10}^{-12}\text{ J }}{\text{7 nucleons}}=9.00\times {10}^{-13}\text{ J/nucleon}\text{.}$

Therefore, the nuclear binding energy per nucleon is $9.00\times {10}^{-13}\text{ J/nucleon}$.

b) ${}_{17}^{35}\text{Cl}$ $\text{34}\text{.96885 amu}$.

In the given nucleus, ${}_{17}^{35}\text{Cl}$, the number of protons and neutrons are $17\text{ and }18$, respectively.

The mass of $17$ ${}_1^1\text{H}$ atoms is as follows:

$\begin{array}{c}{\text{Mass of 17 }}_1^1\text{H }=\text{17}\times \text{mass of hydrogen}\\ =\left(17\times 1.007825\text{ amu}\right)\\ =17.133025\text{ amu}\end{array}$

$\begin{array}{c}\text{Mass of 18 neutrons }=18\times \text{mass of neutron}\\ =\left(\text{18×1}\text{.008665 amu}\right)\\ =\text{18}\text{.15597 amu}\end{array}$

The predicted mass of ${}_{17}^{35}\text{Cl}$ is calculated as follows:

$\begin{array}{c}\text{Predicted mass }={\text{ Mass of 17 }}_1^1\text{H}+\text{mass of 18 neutrons}\\ =\left(17.133025\text{ amu}+\text{18}\text{.15597 amu}\right)\\ =35.28899\text{ amu}\end{array}$

The mass defect is calculated by using the relation as follows:

$\Delta m=\text{Given mass }-\text{Predicted mass}$

Substitute $35.28899\text{ amu}$ for predicted mass and $\text{34}\text{.96885 amu}$ for a given mass,

$\begin{array}{c}\Delta m=\left(\text{34}\text{.96885 amu}\right)-\left(35.28899\text{ amu}\right)\\ =-0.320145\text{ amu}\end{array}$

The value of mass defect is converted into Kg by using the relation:

$\begin{array}{c}1\text{ Kg}=6.0221418\times {10}^{26}\text{ amu}\\ \left(-0.320145\cancel{\text{amu}}\times \frac{1\text{ Kg}}{6.022\times {10}^{26}\cancel{\text{amu}}}\right)=-5.3162\times {10}^{-28}\text{Kg}\end{array}$

The nuclear binding energy is calculated by the relation as follows:

$\Delta E=\left(\Delta m\right){\text{c}}^2$

Substitute $-5.3162\times {10}^{-28}\text{Kg}$ for $\Delta m$ and $3.00\times {10}^8\text{ m/s}$ for c in the above equation,

$\begin{array}{c}\Delta \text{E=}\left(-5.3162\times {10}^{-28}\text{Kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2\\ =-47.8\times {10}^{-12}\text{ Kg}{\text{.m}}^2/{\text{s}}^2\\ =-4.78\times {10}^{-11}\text{ J }\end{array}$

The magnitude of binding energy is considered.

Now, the nuclear binding energy per nucleon is calculated as follows:

$\frac{\text{Nuclear binding energy}}{\text{total number of nucleons}}$

The number of nucleons is defined as the sum of protons and neutrons.

Thus, $35$ nucleons are present in nucleus ${}_{17}^{35}\text{Cl}$.

Substitute $4.78\times {10}^{-11}\text{ J }$ for nuclear binding energy and $35$ nucleons for the total number of nucleons,

$\left(\frac{4.78\times {10}^{-11}\text{ J }}{\text{35 nucleons}}\right)=1.37\times {10}^{-12}\text{ J/nucleon}$

Therefore, the nuclear binding energy per nucleon is $1.37\times {10}^{-12}\text{ J/nucleon}\text{.}$