Chapter 31, Problem 29P

To determine

To find:The energy released per gram of fuel. Compare it to uranium fission energy

Explanation of Solution

Uranium fission released $200\text{MeV}$ per nucleus

The mass of nucleus is 235u

The fusion reactions most likely to succeed in a reactor involve the isotopes of hydrogen, ${}_1^2\text{H}$ (deuterium) and ${}_1^3\text{H}$ (tritium), and are as follows, with the energy released given in parentheses:

  $\begin{array}{ll}{}_1^2\text{H}{+}_1^2\text{H}{\to }_1^3\text{H}{+}_1^1\text{H}\hfill & (4.03\text{MeV})(31-8\mathrm{a})\hfill \\ {}_1^2\text{H}{+}_1^2\text{H}{\to }_2^3\text{He}+\text{n}\hfill & (3.27\text{MeV})(31-8\mathrm{b})\hfill \\ {}_1^2\text{H}{+}_1^3\text{H}{\to }_2^4\text{He}+\text{n}.\hfill & (17.59\text{MeV})(31-8\mathrm{c})\hfill \end{array}$

Eq. 31-8a: ${}_1^2\text{H}{+}_1^2\text{H}{\to }_1^3\text{H}{+}_1^1\text{H}$

  $\begin{array}{l}\frac{4.03\text{MeV}}{2(2.014102\text{u})}\times \frac{1\text{u}}{1.66\times {10}^{-27}\text{kg}}\times \frac{1\text{kg}}{1000\text{g}}\\ =6.03\times {10}^{23}\text{MeV}/\text{g}\end{array}$

Eq. $31-8\text{b}:$

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

  $\begin{array}{l}\frac{3.27\text{MeV}}{2(2.014102\text{u})}\times \frac{1\text{u}}{1.66\times {10}^{-27}\text{kg}}\times \frac{1\text{kg}}{1000\text{g}}\\ =4.89\times {10}^{23}\text{MeV}/\text{g}\end{array}$

Eq. 31-8c: ${}_1^2\text{H}{+}_1^3\text{H}{\to }_2^4\text{He}+\text{n}.$

  $\begin{array}{l}\frac{17.59\text{MeV}}{(2.014102\text{u}+3.016049\text{u})}\times \frac{1\text{u}}{1.66\times {10}^{-27}\text{kg}}\times \frac{1\text{kg}}{1000\text{g}}\\ =2.11\times {10}^{24}\text{MeV}/\text{g}\end{array}$

Energy release per gram of Uranium fission is

  $\frac{200\text{MeV}}{(235\text{u})}\times \frac{1\text{u}}{1.66\times {10}^{-27}\text{kg}}\times \frac{1\text{kg}}{1000\text{g}}=5.13\times {10}^{23}\text{MeV}/\text{g}$

On comparing fuel reaction energy with Uranium fission.

Eq. 31-8a: $\frac{5.13\times {10}^{23}}{6.03\times {10}^{23}}=0.851$

Eq. $31-8\text{b}:\frac{5.13\times {10}^{23}}{4.89\times {10}^{23}}=1.05$

Eq. $31-8\text{c}:\frac{5.13\times {10}^{23}}{2.11\times {10}^{24}}=0.243$