Chapter 20, Problem 20QP

Interpretation Introduction

Interpretation:

The nuclear binding energy and the 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 represented 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 mass defect is as follows:

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

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

Answer to Problem 20QP

Solution:

$4.62\times {10}^{-12}\text{ J }$, $1.15\times {10}^{-12}\text{ J/nucleon}$.

$2.36\times {10}^{-10}\text{ J }$, $1.28\times {10}^{-12}\text{ J/nucleon}$.

Explanation of Solution

a) ${}_2^4\text{He}$ ($\text{4}\text{.002603 amu}$).

In the given nucleus, ${}_2^4\text{He}$, the number of protons and neutrons are $\text{2 and 2}$, respectively.

The mass of $2$ ${}_1^1\text{H}$ atoms is as follows:

$\begin{array}{c}{\text{Mass of 2 }}_1^1\text{H }=\left(2\times \text{mass of hydrogen}\right)\\ =\left(2\times 1.007825\text{ amu}\right)\\ =2.01565\text{ amu}\end{array}$

$\begin{array}{c}\text{Mass of 2 neutrons }=2\times \text{mass of neutron}\\ =\left(\text{2×1}\text{.008665 amu}\right)\\ =\text{2}\text{.01733 amu}\end{array}$

The predicted mass of ${}_2^4\text{He}$ is calculated as follows:

$\begin{array}{c}\text{Predicted mass }={\text{ Mass of 2 }}_1^1\text{H}+\text{Mass of 2 neutrons}\\ =\left(2.01565\text{ amu}+\text{2}\text{.01733 amu}\right)\\ =\text{4}\text{.032980 amu}\end{array}$

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

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

Substitute $\text{4}\text{.032980 amu}$ for predicted mass and $\text{4}\text{.002603 amu}$ for given mass,

$\begin{array}{c}\Delta m=\left(\text{4}\text{.002603 amu}\right)-\left(\text{4}\text{.032980 amu}\right)\\ =-0.03092\text{ amu}\end{array}$

The value of mass defect is converted into Kg by using the relation represented as follows:.

$\begin{array}{c}1\text{ Kg}=6.0221418\times {10}^{26}\text{ amu}\\ \left(-0.03092\cancel{\text{amu}}\times \frac{1\text{ Kg}}{6.022\times {10}^{26}\cancel{\text{amu}}}\right)=-5.134\times {10}^{-29}\text{Kg}\end{array}$

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

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

Substitute $-5.134\times {10}^{-29}\text{Kg}$ for $\Delta \text{m}$ and $3.00\times {10}^8\text{ m/s}$ for c in the equationas follows:

$\begin{array}{c}\Delta E=\left(-5.134\times {10}^{-29}\text{Kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2\\ =-46.2\times {10}^{-13}\text{ Kg}{\text{.m}}^2/{\text{s}}^2\\ =-4.62\times {10}^{-12}\text{ J }\end{array}$

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, $4$ nucleons are present in nucleus ${}_2^4\text{He}$.

Substitute $4.62\times {10}^{-12}\text{ J }$ for nuclear binding energy and $4$ nucleons for the total number of nucleons.

$\left(\frac{4.62\times {10}^{-12}\text{ J }}{\text{4 nucleons}}\right)=1.15\times {10}^{-12}\text{ J/nucleon}$

Therefore, nuclear binding energy per nucleon is $1.15\times {10}^{-12}\text{ J/nucleon}$.

b) ${}_{74}^{184}\text{W}$ ($\text{183}\text{.950928 amu}$)

In the given nucleus, ${}_{74}^{184}\text{W}$, the number of protons and neutrons are $\text{74 and 110}$, respectively.

The mass of $74$ ${}_1^1\text{H}$ atoms is as follows:

$\begin{array}{c}{\text{Mass of 74 }}_1^1\text{H }=\text{74}\times \text{mass of hydrogen}\\ =\left(74\times 1.007825\text{ amu}\right)\\ =\text{74}\text{.57905 amu}\end{array}$

$\begin{array}{c}\text{Mass of 110 neutrons }=110\times \text{mass of neutron}\\ =\left(\text{110×1}\text{.008665 amu}\right)\\ =\text{110}\text{.95315 amu}\end{array}$

The predicted mass of ${}_{74}^{184}\text{W}$ is calculated as follows:

$\begin{array}{c}\text{Predicted mass }={\text{ Mass of 74 }}_1^1\text{H}+\text{Mass of 110 neutrons}\\ =\left(\text{74}\text{.57905 amu}+\text{110}\text{.95315 amu}\right)\\ =185.5323\text{ amu}\end{array}$

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

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

Substitute $185.5323\text{ amu}$ for predicted mass and $\text{183}\text{.950928 amu}$ for given mass,

$\begin{array}{c}\Delta \text{m}=\left(\text{183}\text{.950928 amu}\right)-\left(185.5323\text{ amu}\right)\\ =-1.5814\text{ amu}\text{.}\end{array}$

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

$\begin{array}{c}1\text{ Kg}=6.0221418\times {10}^{26}\text{ amu}\\ \left(-1.5814\cancel{\text{amu}}\times \frac{1\text{ Kg}}{6.022\times {10}^{26}\cancel{\text{amu}}}\right)=-2.626\times {10}^{-27}\text{Kg}\end{array}$

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

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

Substitute $-2.626\times {10}^{-27}\text{Kg}$ for $\Delta \text{m}$ and $3.00\times {10}^8\text{ m/s}$ for c in the above expression,

$\begin{array}{c}\Delta E=\left(-2.626\times {10}^{-27}\text{Kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2\\ =-2.36\times {10}^{-10}\text{ Kg}{\text{.m}}^2/{\text{s}}^2\\ =-2.36\times {10}^{-10}\text{ J }\end{array}$

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, $184$ nucleons are present in nucleus ${}_{74}^{184}\text{W}$.

Substitute $2.36\times {10}^{-10}\text{ J }$ for nuclear binding energy and $184$ nucleons for thetotal number of nucleons,

$\left(\frac{2.36\times {10}^{-10}\text{ J }}{\text{184 nucleons}}\right)=1.28\times {10}^{-12}\text{ J/nucleon}$

Therefore, the nuclear binding energy per nucleon is $1.28\times {10}^{-12}\text{ J/nucleon}$.