Chapter 27, Problem 43QAP

To determine

(a)

The binding energy of per nucleon of $H$

Explanation of Solution

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $H=2.01355321270u$

  $H+Energy\to p^{+}+n^0$

Mass defect

  $\begin{array}{l}\Delta m=\left(1.007276467u+1.008664917u\right)-2.01355321270\\ =0.023881713u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.023881713u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =2.224567MeV\end{array}$

The $H$ nucleus has four nucleons (one neutron and one proton), therefore the binding energy per nucleon is

  $\frac{E}{2}=\frac{2.224567MeV}{2}=\text{1}\text{.1122835}\text{MeV/nucleon}$

Conclusion:

The binding energy of per nucleon of $H$ is $\text{1}\text{.1122835}\text{MeV/nucleon}$.

To determine

(b)

The binding energy of per nucleon of $He$

Explanation of Solution

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $He=4.002602u$

  $He+Energy\to 2p^{+}+2n^0$

Mass defect

  $\Delta m=(2\times 1.008665+2\times 1.007825)-4.002602=0.030378u$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.030378u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =28.2992659MeV\end{array}$

The $He$ nucleus has four nucleons (two neutrons and one protons), therefore the binding energy per nucleon is

  $\frac{E}{4}=\frac{28.2992659MeV}{4}=7.0742\text{MeV/nucleon}$

Conclusion:

The binding energy of per nucleon of $He$ is $7.0742\text{MeV/nucleon}$.

To determine

(c)

The binding energy of per nucleon of $Li$

Explanation of Solution

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $Li=6.01348u$

  $Li+Energy\to 3p^{+}+4n^0$

Mass defect

  $\begin{array}{l}\Delta m=(3\times 1.008664917+3\times 1.007276467)-6.01348\\ =0.034344152u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.034344152u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =31.99137362MeV\end{array}$

The $Li$ nucleus has 6 nucleons therefore the binding energy per nucleon is

  $\frac{E}{6}=\frac{31.99137362MeV}{6}=5.331\text{MeV/nucleon}$

Conclusion:

The binding energy of per nucleon of $Li$ is $5.331\text{MeV/nucleon}$.

To determine

(d)

The binding energy of per nucleon of $C$

Explanation of Solution

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $C=12.000u$

  $C+Energy\to 6p^{+}+6n^0$

Mass defect

  $\begin{array}{l}\Delta m=(6\times 1.008664917+6\times 1.007276467)-12\\ =0.095648304u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.095648304u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =89.0958MeV\end{array}$

The $C$ nucleus has 12 nucleons therefore the binding energy per nucleon is

  $\frac{E}{12}=\frac{89.0958MeV}{12}=7.6802\text{MeV/nucleon}$

Conclusion:

The binding energy of per nucleon of $C$ is $7.6802\text{MeV/nucleon}$.

To determine

(e)

The binding energy of per nucleon of $Fe$

Explanation of Solution

Calculation:

$$

There are 26 protons and 30 neutrons in the stable iron isotope (Iron / Fe-56). Isotopes are atoms of the same element with the same number of atoms (the same number of protons), but with a different mass. The atomic number of all iron isotopes is 26. However, their masses are different.

Natural iron (Fe) consists of 4 isotopes: 5.845% of radioactive 54Fe, 91.754% of stable 56Fe, 2.119% of stable 57Fe and 0.282% of stable 58Fe. 56Fe of which is the most common because it is the most common endpoint of fusion chains in extremely massive stars.

The mass of Iron/Fe-56 $M(Fe)=55.9349375u$

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

The weight of an electron (it has a weight of about 1/1836 that of the proton):

  $M_e=5.4857990943\times {10}^{-4}u$

  $Fe+Energy\to 26p^{+}+30n^0$

  $26M(H)+30m_n=56.46339748u$

  $\Delta m=(56.46339748u-55.9349375u)=0.52845998u$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.52845998u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =492.26MeV\end{array}$

The $Fe$ nucleus has 56 nucleons therefore the binding energy per nucleon is

  $\frac{E}{56}=\frac{492.26MeV}{56}=8.79031\text{MeV/nucleon}$

Conclusion:

The binding energy of per nucleon of $Fe$ is $8.79031\text{MeV/nucleon}$.

To determine

(f)

The binding energy of per nucleon of $Sr$

Explanation of Solution

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $Sr=89.908u$

  $Sr+Energy\to 38p^{+}+52n^0$

Mass defect

  $\begin{array}{l}\Delta m=(52\times 1.008664917+38\times 1.007276467)-89.908\\ =1.377170663u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=1.377170663u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =1282.826294MeV\end{array}$

The $Sr$ nucleus has 90 nucleons therefore the binding energy per nucleon is

  $\frac{E}{90}=\frac{1282.826294MeV}{90}=8.69594\text{MeV/nucleon}$

Conclusion:

The binding energy of per nucleon of $Sr$ is $8.69594\text{MeV/nucleon}$.

To determine

(g)

The binding energy of per nucleon of $I$

Explanation of Solution

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $I=89.908u$

  $I+Energy\to 53p^{+}+76n^0$

Mass defect

  $\begin{array}{l}\Delta m=(76\times 1.008664917+53\times 1.007276467)-129.913\\ =0.11186443u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.13186443u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =121.1994MeV\end{array}$

The $I$ nucleus has 129 nucleons therefore the binding energy per nucleon is

  $\frac{E}{129}=\frac{121.1994MeV}{129}=8.43602\text{MeV/nucleon}$

Conclusion:

The binding energy of per nucleon of $U$ is $8.43602\text{MeV/nucleon}$.

To determine

(H)

The binding energy of per nucleon of $U$

Explanation of Solution

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $U=235.04393$

  $I+Energy\to 53p^{+}+76n^0$

Mass defect

  $\begin{array}{l}\Delta m=(143\times 1.008664917+92\times 1.007276467)-235.04393\\ =1.864588095u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=1.864588095u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =\text{1736}\text{.852736703803795}\text{MeV}\end{array}$

The $U$ nucleus has 235 nucleons therefore the binding energy per nucleon is

  $\frac{E}{235}=\frac{\text{1736}\text{.852736703803795}\text{MeV}}{235}=7.39086\text{MeV}/\text{nucleon}$

Conclusion:

The binding energy of per nucleon of $U$ is $7.39086\text{MeV}/\text{nucleon}$.