Chapter 29, Problem 64P

To determine

The comparison of the amount of energy released when $1.0\text{ kg}$ of ${}^{235}\text{U}$ undergoes the fission reaction with the energy released when $1.0\text{ kg}$ of hydrogen undergoes the fusion reaction.

Answer to Problem 64P

The fission reaction of ${}^{235}\text{U}$ releases $7.1\times {10}^{13}\text{ J/kg}$ whereas the fusion reaction of hydrogen releases $34\times {10}^{13}\text{ J/kg}$ .

Explanation of Solution

Write the equation for the energy released.

  $E=\left|\Delta m\right|c^2$        (I)

Here, $E$ is the energy released, $\Delta m$ is the change in mass during the reaction and $c$ is the velocity of light.

Write the chemical equation for the fission reaction of ${}^{235}\text{U}$ nucleus.

  ${}_0^1\text{n}{+}_{92}^{235}\text{U}{\to }_{56}^{141}\text{Ba}{+}_{36}^{92}\text{Kr}+3_0^1\text{n}$

Write the equation for $\Delta m$ using the chemical equation of the fission reaction of ${}^{235}\text{U}$ .

  $\Delta m=m_{\text{n}}+m_{\text{U}}-\left(m_{\text{Ba}}+m_{\text{Kr}}+3m_{\text{n}}\right)$        (II)

Here, $m_{\text{n}}$ is the mass of neutron, $m_{\text{U}}$ is the mass of ${}^{235}\text{U}$ nucleus, $m_{\text{Ba}}$ is the mass of ${}_{56}^{141}\text{Ba}$ nucleus and $m_{\text{Kr}}$ is the mass of ${}_{36}^{92}\text{Kr}$ nucleus.

Write the chemical equation for the fusion reaction of hydrogen.

  ${}_1^2\text{H}{+}_1^3\text{H}{\to }_2^4\text{He}{+}_0^1\text{n}$

Write the equation for $\Delta m$ using the chemical equation of the fusion reaction of hydrogen.

  $\Delta m=m_{{}^2\text{H}}+m_{{}^3\text{H}}-\left(m_{\text{He}}+m_{\text{n}}\right)$        (III)

Here, $m_{{}^2\text{H}}$ is the mass of ${}_1^2\text{H}$ nucleus, $m_{{}^3\text{H}}$ is the mass of ${}_1^3\text{H}$ nucleus and $m_{\text{He}}$ is the mass of ${}_2^4\text{He}$ nucleus.

Write the equation for the total energy released from $1.0\text{ kg}$ of ${}^{235}\text{U}$ .

Conclusion:

The value of $m_n$ is $1.0086649\text{ u}$ , that of $m_{\text{U}}$ is $235.0439282\text{ u}$ , that of $m_{\text{Ba}}$ is $140.9144035\text{ u}$ , that of $m_{\text{Kr}}$ is $91.9261731\text{ u}$ , that of $m_{{}^2\text{H}}$ is $2.0141018\text{ u}$ , that of $m_{{}^3\text{H}}$ is $3.0160493\text{ u}$ , that of $m_{\text{He}}$ is $4.0026033\text{ u}$ and that of $c^2$ is $931.494\text{ MeV/u}$ .

Substitute $1.0086649\text{ u}$ for $m_n$ , $235.0439282\text{ u}$ for $m_{\text{U}}$ , $140.9144035\text{ u}$ for $m_{\text{Ba}}$ and $91.9261731\text{ u}$ for $m_{\text{Kr}}$ in equation (II) to find $\Delta m$ of the fission reaction of ${}^{235}\text{U}$ .

  $\begin{array}{c}\Delta m=1.0086649\text{ u}+235.0439282\text{ u}-\left(140.9144035\text{ u}+91.9261731\text{ u}+3\left(1.0086649\text{ u}\right)\right)\\ =0.1860218\text{ u}\end{array}$

Substitute $0.1860218\text{ u}$ for $\Delta m$ and $931.494\text{ MeV/u}$ for $c^2$ in equation (I) to find $E$ for the fission reaction of ${}^{235}\text{U}$.

  $\begin{array}{c}{E}_{\text{fission}}=\left|0.1860218\text{ u}\right|\left(931.494\text{ MeV/u}\right)\\ =173.278\text{ MeV}\end{array}$

Write the equation for the total energy released for $1.0\text{ kg}$ of ${}^{235}\text{U}$ .

  $E_{\text{fission,tot}}=E_{\text{fission}}\frac{1.0\text{ kg}}{m_{\text{U}}}$

Here, $E_{\text{fission,tot}}$ is total energy released for $1.0\text{ kg}$ of ${}^{235}\text{U}$ .

Substitute $173.278\text{ MeV}$ for $E_{\text{fission}}$ and $235.0439282\text{ u}$ for $m_{\text{U}}$ in the above equation to find $E_{\text{fission,tot}}$ .

  $\begin{array}{c}{E}_{\text{fission,tot}}=\left(173.278\text{ MeV}\frac{{10}^6\text{ eV}}{1\text{ MeV}}\right)\frac{1.0\text{ kg}}{\left(235.0439282\text{ u}\frac{1.6605\times {10}^{-27}\text{ kg}}{1\text{ u}}\right)}\\ =4.44\times {10}^{32}\text{ eV}\frac{1.602\times {10}^{-19}\text{ J}}{1\text{ eV}}\\ =7.1\times {10}^{13}\text{ J}\end{array}$

Substitute $2.0141018\text{ u}$ for $m_{{}^2\text{H}}$ , $3.0160493\text{ u}$ for $m_{{}^3\text{H}}$ , $4.0026033\text{ u}$ for $m_{\text{He}}$ and $1.0086649\text{ u}$ for $m_n$ in equation (III) to find $\Delta m$ of the fusion reaction of hydrogen.

  $\begin{array}{c}\Delta m=2.0141018\text{ u}+3.0160493\text{ u}-\left(4.0026033\text{ u}+1.0086649\text{ u}\right)\\ =0.0188829\text{ u}\end{array}$

Substitute $0.0188829\text{ u}$ for $\Delta m$ and $931.494\text{ MeV/u}$ for $c^2$ in equation (I) to find $E$ for the fusion reaction of hydrogen.

  $\begin{array}{c}{E}_{\text{fusion}}=\left|0.0188829\text{ u}\right|\left(931.494\text{ MeV/u}\right)\\ =17.5893\text{ MeV}\end{array}$

Write the equation for the total energy released from $1.0\text{ kg}$ of hydrogen.

  $E_{\text{fusion,tot}}=E_{\text{fusion}}\frac{1.0\text{ kg}}{m_{{}^2\text{H}}+m_{{}^3\text{H}}}$

Here, $E_{\text{fusion,tot}}$ is total energy released from $1.0\text{ kg}$ of hydrogen.

Substitute $17.5893\text{ MeV}$ for $E_{\text{fusion}}$ , $2.0141018\text{ u}$ for $m_{{}^2\text{H}}$ , $3.0160493\text{ u}$ for $m_{{}^3\text{H}}$ in the above equation to find $E_{\text{fission,tot}}$ .

$\begin{array}{c}{E}_{\text{fusion,tot}}=\left(17.5893\text{ MeV}\frac{{10}^6\text{ eV}}{1\text{ MeV}}\right)\frac{1.0\text{ kg}}{\left(2.0141018\text{ u}\frac{1.6605\times {10}^{-27}\text{ kg}}{1\text{ u}}+3.0160493\text{ u}\frac{1.6605\times {10}^{-27}\text{ kg}}{1\text{ u}}\right)}\\ =2.11\times {10}^{33}\text{ eV}\frac{1.602\times {10}^{-19}\text{ J}}{1\text{ eV}}\\ =34\times {10}^{13}\text{ J}\end{array}$

Therefore, the fission reaction of ${}^{235}\text{U}$ releases $7.1\times {10}^{13}\text{ J/kg}$ whereas the fusion reaction of hydrogen releases $34\times {10}^{13}\text{ J/kg}$ .