Chapter 31, Problem 34P

To determine

To find: Energy obtain from natural deuterium of water used in fusion reaction, and to compare the energy with the gasoline.

Answer to Problem 34P

  $2.46\times {10}^9\text{J}$ ; 50 times more than gasoline.

Explanation of Solution

Given data:

(In equation 31-8a,) Energy released for every 2 deuterium atoms is $4.00\text{MeV}$

Energy released from 1.0 kg gasoline, ${\mathrm{E}}_{\text{gasoline}}=5\times {10}^7\text{J}$

Calculation:

Number of natural deuterium nuclei in 1.00 kg of water.$\left(1.00{\text{kgH}}_2\text{O}\right)\left(\frac{6.02\times {10}^{23}{\text{H}}_2\text{O}}{0.018{\text{kgH}}_2\text{O}}\right)\left(\frac{2\text{H}}{1{\text{H}}_2\text{O}}\right)\left(\frac{1.15\times {10}^{-4}\text{d}}{1\text{H}}\right)=7.692\times {10}^{21}\text{d}$ nuclei

Energy obtain from natural deuterium of water used in fusion reaction

  $(7.692\times {10}^{21}\text{d}$ nuclei $)\left(\frac{4.00\times {10}^6\text{eV}}{2\text{d atoms }}\right)\left(\frac{1.60\times {10}^{-19}\text{J}}{1\text{eV}}\right)=2.46\times {10}^9\text{J}$

As compared to gasoline,

Fusion reaction energy is 50 times more than energy $\frac{2.46\times {10}^9\text{J}}{5\times {10}^7\text{J}}\approx 50$

So fusion reaction energy is 50 times more than gasoline.

Conclusion:

Energy obtain from natural deuterium of water used in fusion reaction is $2.46\times {10}^9\text{J}$ . It is 50 times more than gasoline.