Question

Consider equal numbers of deuterium and tritium nuclei fusing to form helium-4 nuclei, as given in equation,

(a) What is the yield in joules of kinetic energy liberated per kilogram of fuel?

(b) How does this compare with a typical yield of${\text{10}}^{\text{6}}\text{J/kg}$ for chemical fuels?

Short answer

1. $3.4\times {10}^{14}\text{J}$
   
2. Calculated kinetic energy is $3.4\times {10}^8$multiple of the kinetic energy of chemical fuels.

Step-by-step solution

Given data and concept

Given reaction is:

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

The molecular mass of D is $M_D=2.01414u$,

The molecular mass of T is $M_T=3.0160u$,

The energy released in this decay depends upon the mass difference between the parent and daughter nucleus.

We can calculate the kinetic energy Q of the emitted particle using the formula:

role="math" localid="1658414034727" $Q=\left(m_p-m_D\right)c^2$

In this formula ${\text{m}}_P$is the mass of the parent nucleus. And${\text{m}}_D$ is the mass of the daughter nucleus.

(a) Kinetic energy

The energy released from the above reaction is:

$Q=17.6MeV=17.6\times 1.6\times {10}^{-13}J=2.819\times {10}^{-12}J$

Now for the sum of molecular masses of deuterium and tritium:

$\begin{array}{rcl}M& =& M_D+M_T\\ & =& 2.0141u+3.016u\\ & =& 5.0301u\\ & =& 8.35\times {10}^{-27}kg\\ & & \end{array}$

The number of pairs of deuterium and tritium nuclei in 1kg of fuel:

$\begin{array}{rcl}N& =& \frac{m}{M}\\ & =& \frac{1kg}{8.35\times {10}^{-27}kg}\\ & =& 0.12\times {10}^{27}\\ & & \end{array}$

The energy released per kilogram of fuel is:

$\begin{array}{rcl}KE& =& Q\times N\\ & =& 2.819\times {10}^{-12}J\times 0.12\times {10}^{27}\\ & =& 3.4\times {10}^{14}J\\ & & \end{array}$

Therefore, kinetic energy per 1 kg is$KE=3.4\times {10}^{14}J$

(b) Comparision

Given kinetic energy for chemical fuels is ${10}^6\text{J}$.

The calculated energy is$3.4\times {10}^{14}J$ .

A comparison of the above two can be calculated as:

$$\begin{gathered} =\frac{3.4\times {10}^{14}J}{{10}^6\text{J}} \\ =3.4\times {10}^8 \end{gathered}$$

Thus, calculated kinetic energy is $3.4\times {10}^8$multiple of the kinetic energy of chemical fuels.