Chapter 30.2, Problem 34SSC

To determine

The energy released in the first fusion.

Answer to Problem 34SSC

The energy released in the first fusion is $1.4928\times {10}^{-13}\text{ J}$ .

Explanation of Solution

Given:

Mass of $\text{H}$ , $m_{\text{H}}=1.007825\text{ u}$

Mass of $\text{H}$ , $m_{\text{H}}=2.014102\text{ u}$

Mass of $\text{e}$ , $m_{\text{e}}=0.000548756\text{ u}$

Formula used:

  $E=\Delta m{c}^2$

Where, $\Delta m$ = mass defect

  $c$ = speed of light

Calculation:

The mass defect is calculated as,

  $\begin{array}{l}\Delta m=2\left(m_{\text{H}}\right)-m_{\text{e}}-m_{\text{H}}\\ \Delta m=2\left(1.007825\text{ u}\right)-\left(0.000548756\text{ u}\right)-2.014102\text{ u}\\ \Delta m=9.99244\times {10}^{-4}\text{ u}\end{array}$

Now, converting mass defect from atomic mass unit to kg,

  $\text{1 u =1}\text{.66}\times {\text{10}}^{-27}\text{ kg}$

  $\begin{array}{l}\Delta m=\left(9.99244\times {10}^{-4}\text{ u}\right)\left(\text{1}\text{.66}\times {\text{10}}^{-27}\frac{\text{kg}}{\text{u}}\right)\\ \Delta m=1.6587\times {10}^{30}\text{ kg}\end{array}$

The total energy released is calculated as,

  $\begin{array}{l}E=\Delta m{c}^2\\ E=\left(1.6587\times {10}^{-30}\right){\left(3\times {10}^8\right)}^2\\ E=1.4928\times {10}^{-13}\text{kg}\end{array}$

Conclusion:

  $E=1.4928\times {10}^{-13}\text{ kg}$.