Chapter 20, Problem 68AP

Interpretation Introduction

Interpretation:

The nuclear binding energy per nucleon for the given nucleus is to be calculated.

Concept introduction:

The nuclear binding energy is calculated by using the following relation:

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

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

Mass defect is defined as the difference between the given mass and predicted mass. Thus, the formula for the mass defect is as follows:

$\Delta \text{m}=\text{ given mass }-\text{ predicted mass}$

Here, $\Delta \text{m}$ is the mass defect.

Answer to Problem 68AP

Solution:

$1.040\times {10}^{-12}\text{ J/nucleon}$.

$1.111\times {10}^{-12}\text{ J/nucleon}$.

$1.199\times {10}^{-12}\text{ J/nucleon}$.

$1.410\times {10}^{-12}\text{ J/nucleon}$.

Explanation of Solution

a)The mass of ${}_5^{10}\text{B}$ is $\text{10}\text{.0129 amu}$.

In the given nucleus, ${}_5^{10}\text{B}$, the number of protons and neutrons is 5 each.

The mass of $5$ ${}_1^1\text{H}$ atoms is as follows:

$\begin{array}{l}{\text{Mass of 5 }}_1^1\text{H =5}\times \text{mass of hydrogen}\\ =5\times 1.007825\text{ amu}\\ =5.039125\text{ amu}\text{.}\end{array}$

$\begin{array}{l}\text{Mass of 5 neutrons }=5\times \text{mass of neutron}\\ \text{=5×1}\text{.008665 amu}\\ \text{=5}\text{.043325 amu}\text{.}\end{array}$

The predicted mass of ${}_5^{10}\text{B}$ is iscalculated as follows:

$\begin{array}{c}{\text{Predicted mass = mass of 5 }}_1^1\text{H}+\text{mass of 5 neutrons}\\ =5.039125\text{ amu}+\text{5}\text{.043325 amu}\\ =10.08245\text{ amu}\end{array}$

The mass defect is calculated by using the relation given below:

$\Delta \text{m}=\text{ given mass }-\text{ predicted mass}$

Here, $\Delta \text{m}$ is the mass defect.

Substitute $10.08245\text{ amu}$ for the predicted mass and $\text{10}\text{.0129 amu}$ for the given mass.

$\begin{array}{c}\Delta \text{m}=\left(\text{10}\text{.0129 amu}\right)-\left(10.08245\text{ amu}\right)\\ =0.06955\text{amu}\text{.}\end{array}$

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

$\begin{array}{l}1\text{ kg}=6.0221418\times {10}^{26}\text{ amu}\\ \left(0.06955\cancel{\text{amu}}\times \frac{1\text{ kg}}{6.022\times {10}^{26}\cancel{\text{amu}}}\right)=1.1549\times {10}^{-28}\text{kg}\end{array}$

The nuclear binding energy is calculated by the relation given below:

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

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

Substitute $1.1549\times {10}^{-28}\text{Kg}$ for $\Delta \text{m}$ and $3.00\times {10}^8\text{ m/s}$ for c in the above equation as follows:

$\begin{array}{c}\Delta \text{E=}\left(1.1549\times {10}^{-28}\text{Kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2\\ =10.39\times {10}^{-12}\text{ Kg}{\text{.m}}^2/{\text{s}}^2\\ =1.039\times {10}^{-11}\text{ J }\end{array}$

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 the protons and neutrons.

Thus, $10$ nucleons are present in the nucleus of ${}_5^{10}\text{B}$.

Substitute $1.039\times {10}^{-11}\text{ J }$ for the nuclear binding energy and $10$ nucleons for the total number of nucleons,

$\frac{1.039\times {10}^{-11}\text{ J }}{\text{10 nucleons}}=1.039\times {10}^{-12}\text{ J/nucleon}\cong 1.040\times {10}^{-12}\text{ J/nucleon}$

Therefore, the nuclear binding energy per nucleon is $1.040\times {10}^{-12}\text{ J/nucleon}$

b)The mass of ${}_5^{11}\text{B}$ is $\text{11}\text{.009305 amu}$.

In the given nucleus, ${}_5^{11}\text{B}$, the number of protons and neutrons are $\text{5 and 6}$, respectively.

The mass of $5$ ${}_1^1\text{H}$ atoms is calculated as follows:

$\begin{array}{l}{\text{Mass of 5 }}_1^1\text{H =5}\times \text{mass of hydrogen}\\ =5\times 1.007825\text{ amu}\\ =5.039125\text{ amu}\text{.}\end{array}$

$\begin{array}{l}\text{Mass of 6 neutrons }=6\times \text{mass of neutron}\\ \text{=6×1}\text{.008665 amu}\\ \text{=6}\text{.05199 amu}\text{.}\end{array}$

The predicted mass of ${}_5^{11}\text{B}$ is calculated as follows:

$\begin{array}{c}{\text{Predicted mass = mass of 5 }}_1^1\text{H}+\text{mass of 6 neutrons}\\ =5.039125\text{ amu}+\text{6}\text{.05199 amu}\text{.}\\ =11.09111\text{ amu}\end{array}$

The mass defect is calculated by using the relation given below:

$\Delta \text{m}=\text{ given mass - predicted mass}$

Here, $\Delta \text{m}$ is the mass defect.

Substitute $11.09111\text{ amu}$ for the predicted mass and $\text{11}\text{.009305 amu}$ for the given mass,

$\begin{array}{c}\Delta \text{m}=\left(\text{11}\text{.009305 amu}\right)-\left(11.09111\text{ amu}\right)\\ =0.08181\text{amu}\text{.}\end{array}$

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

$\begin{array}{l}1\text{ Kg}=6.0221418\times {10}^{26}\text{ amu}\\ \left(0.08181\cancel{\text{amu}}\times \frac{1\text{ Kg}}{6.022\times {10}^{26}\cancel{\text{amu}}}\right)=1.3585\times {10}^{-28}\text{Kg}\end{array}$

The nuclear binding energy is calculated by the relation given below:

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

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

Substitute $1.3585\times {10}^{-28}\text{Kg}$ for $\Delta \text{m}$ and $3.00\times {10}^8\text{ m/s}$ for c in the above equation as follows:

$\begin{array}{c}\Delta \text{E=}\left(1.3585\times {10}^{-28}\text{Kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2\\ =12.226\times {10}^{-12}\text{ Kg}{\text{.m}}^2/{\text{s}}^2\\ =1.2226\times {10}^{-11}\text{ J }\end{array}$

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 the protons and neutrons.

Thus, $11$ nucleons are present in the nucleus of ${}_5^{11}\text{B}$.

Substitute $1.2226\times {10}^{-11}\text{ J }$ for the nuclear binding energy and $11$ nucleons for the total number of nucleons as:

$\frac{1.2226\times {10}^{-11}\text{ J }}{\text{11 nucleons}}=1.111\times {10}^{-12}\text{ J/nucleon}$

Therefore, the nuclear binding energy per nucleon is $1.111\times {10}^{-12}\text{ J/nucleon}$

c)The mass of ${}_7^{14}\text{N}$ is $\text{14}\text{.003074 amu}$.

In the given nucleus, ${}_7^{14}\text{N}$, the number of protons and neutrons are $\text{7 and 7}$, respectively.

The mass of $7$ ${}_1^1\text{H}$ atoms is calculated as follows:

$\begin{array}{l}{\text{Mass of 7 }}_1^1\text{H =7}\times \text{mass of hydrogen}\\ =7\times 1.007825\text{ amu}\\ =7.05477\text{ amu}\text{.}\end{array}$

$\begin{array}{l}\text{Mass of 7 neutrons }=7\times \text{mass of neutron}\\ \text{=7×1}\text{.008665 amu}\\ \text{=7}\text{.060655 amu}\text{.}\end{array}$

The predicted mass of ${}_7^{14}\text{N}$ is calculated as follows:

$\begin{array}{c}{\text{Predicted mass = mass of 7 }}_1^1\text{H}+\text{mass of 7 neutrons}\\ =7.05477\text{ amu}+\text{7}\text{.060655 amu}\text{.}\\ =14.11542\text{ amu}\end{array}$

The mass defect is calculated by using the relation given below:

$\Delta \text{m}=\text{ given mass }-\text{ predicted mass}$

Here, $\Delta \text{m}$ is the mass defect.

Substitute $14.11542\text{ amu}$ for predicted mass and $\text{14}\text{.003074 amu}$ for given mass as:

$\begin{array}{c}\Delta \text{m}=\left(\text{14}\text{.003074 amu}\right)-\left(14.11542\text{ amu}\right)\\ =0.112346\text{ amu}\text{.}\end{array}$

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

$\begin{array}{l}1\text{ Kg}=6.0221418\times {10}^{26}\text{ amu}\\ \left(0.112346\cancel{\text{amu}}\times \frac{1\text{ Kg}}{6.022\times {10}^{26}\cancel{\text{amu}}}\right)=1.8655\times {10}^{-28}\text{Kg}\end{array}$

The nuclear binding energy is calculated by the relation given below:

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

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

Substitute $1.8655\times {10}^{-28}\text{Kg}$ for $\Delta \text{m}$ and $3.00\times {10}^8\text{ m/s}$ for c in the above equation as follows:

$\begin{array}{c}\Delta \text{E=}\left(1.8655\times {10}^{-28}\text{Kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2\\ =16.789\times {10}^{-12}\text{ Kg}{\text{.m}}^2/{\text{s}}^2\\ =1.6789\times {10}^{-11}\text{ J }\end{array}$

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 the protons and neutrons.

Thus, $14$ nucleons are present in the nucleus of ${}_7^{14}\text{N}$.

Substitute $1.6789\times {10}^{-11}\text{ J }$ for the nuclear binding energy and $14$ nucleons for the total number of nucleons as:

$\frac{1.6789\times {10}^{-11}\text{ J }}{\text{14 nucleons}}=1.199\times {10}^{-12}\text{ J/nucleon}$

Therefore, the nuclear binding energy per nucleon is $1.199\times {10}^{-12}\text{ J/nucleon}$

d)The mass of ${}_{26}^{56}\text{Fe}$ is $\text{55}\text{.93494 amu}$.

In the given nucleus, ${}_{26}^{56}\text{Fe}$, the number of protons and neutrons are $\text{26 and 30}$, respectively.

The mass of $26$ ${}_1^1\text{H}$ atoms is as follows:

$\begin{array}{l}{\text{Mass of 26 }}_1^1\text{H =26}\times \text{mass of hydrogen}\\ =26\times 1.007825\text{ amu}\\ =26.20345\text{ amu}\text{.}\end{array}$

$\begin{array}{l}\text{Mass of 30 neutrons }=30\times \text{mass of neutron}\\ \text{=30×1}\text{.008665 amu}\\ \text{=30}\text{.25995 amu}\text{.}\end{array}$

The predicted mass of ${}_{26}^{56}\text{Fe}$ is calculated as follows:

$\begin{array}{c}{\text{Predicted mass = mass of 26 }}_1^1\text{H}+\text{mass of 30 neutrons}\\ =26.20345\text{ amu}+\text{30}\text{.25995 amu}\text{.}\\ =56.4634\text{ amu}\end{array}$

The mass defect is calculated by using the relation given below:

$\Delta \text{m}=\text{ given mass }-\text{ predicted mass}$

Here, $\Delta \text{m}$ is the mass defect.

Substitute $56.4634\text{ amu}$ for the predicted mass and $\text{55}\text{.93494 amu}$ for the given mass.

$\begin{array}{c}\Delta \text{m}=\left(\text{55}\text{.93494 amu}\right)-\left(56.4634\text{ amu}\right)\\ =0.52846\text{ amu}\text{.}\end{array}$

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

$\begin{array}{l}1\text{ Kg}=6.0221418\times {10}^{26}\text{ amu}\\ \left(0.52846\text{ amu}\text{. }\cancel{\text{amu}}\times \frac{1\text{ Kg}}{6.022\times {10}^{26}\cancel{\text{amu}}}\right)=8.7754\times {10}^{-28}\text{Kg}\end{array}$

The nuclear binding energy is calculated by the relation given below:

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

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

Substitute $8.7754\times {10}^{-28}\text{Kg}$ for $\Delta \text{m}$ and $3.00\times {10}^8\text{ m/s}$ for c in the above equation as follows:

$\begin{array}{c}\Delta \text{E=}\left(8.7754\times {10}^{-28}\text{Kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2\\ =78.9786\times {10}^{-12}\text{ Kg}{\text{.m}}^2/{\text{s}}^2\\ =7.89786\times {10}^{-11}\text{ J }\end{array}$

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 the protons and neutrons.

Thus, $56$ nucleons are present in the nucleus of ${}_{26}^{56}\text{Fe}$.

Substitute $7.89786\times {10}^{-11}\text{ J }$ for the nuclear binding energy and $56$ nucleons for the total number of nucleons,

$\frac{7.89786\times {10}^{-11}\text{ J }}{\text{56 nucleons}}=1.410\times {10}^{-12}\text{ J/nucleon}$

Therefore, the nuclear binding energy per nucleon is $1.410\times {10}^{-12}\text{ J/nucleon}$.