Figure 43-15 shows an early proposal for a hydrogen bomb. The fusion fuel is deuterium,${}^{\mathbf{2}}\mathbf{H}$. The high temperature and particle density needed for fusion are provided by an atomic bomb “trigger” that involves a ${}^{\mathbf{235}}\mathbf{U}$or${}^{\mathbf{239}}\mathbf{Pu}$fission fuel arranged to impress an imploding, compressive shock wave on the deuterium. The fusion reaction is

${\mathbf{5}}^{\mathbf{2}}\mathbf{H}{\mathbf{\to }}^{\mathbf{3}}\mathbf{He}{\mathbf{+}}^{\mathbf{4}}\mathbf{He}{\mathbf{+}}^{\mathbf{1}}\mathbf{H}\mathbf{+}\mathbf{2}\mathbf{n}$

(a) Calculate Q for the fusion reaction. For needed atomic masses, see Problem 42. (b) Calculate the rating (see Problem 16) of the fusion part of the bomb if it contains 500 kg of deuterium, 30.0% of which undergoes fusion.

The expression for energy is given by,

$\mathit{Q}\mathbf{=}\mathbf{-}\mathbf{∆}\mathit{m}{\mathit{c}}^{\mathbf{2}}$

Here,$\mathit{Q}$ is the energy released in a reaction, $\mathbf{∆}\mathit{m}$is the mass difference between the parent nuclei and the daughter nuclei, and $\mathit{c}$is the velocity of light.

Rewrite the energy equation as follows.

$Q=-\left(5m_{2H}-m_{3He}-m_{4He}+m_{1H}-2m_n\right)c^2.....\left(2\right)$

Substitute all the known values in equation (2).

$$\begin{gathered} Q=\left[5\left(2.014102u\right)-3.016029u-4.002603u-1.007825u-2\left(1.008665u\right)\right]\left(931.5MeV/u\right) \\ =\left(0.026723u\right)\left(931.5MeV/u\right) \\ =24.9MeV \end{gathered}$$

Therefore, the value of$Q$is $24.9MeV$.

The number of five deuteriums in this mass equals the number of moles multiplied by the number of atoms in one mole, where the number of moles equals the mass divided by five the molar mass of the deuterium.

$N=\frac{m_{2H}{N}_A}{5M_{2H}}$

Let$Q$ be the energy release per fusion, then the total energy released by fusion equals the number of fusions multiplied by$N$ .

$$\begin{gathered} E_{fusion}=0.3NQ \\ =\frac{0.3m_{2H}{N}_{A}Q}{5M_{2H}} \end{gathered}$$

The expression to find rating is given by,

$$\begin{gathered} R=\frac{E_{fusion}}{2.6\times {10}^{28}MeV/megatonTNT} \\ =\frac{0.3m_{2H}{N}_{A}Q}{5M_{2H}\left(2.6\times {10}^{28}MeV/megatonTNT\right)}....\left(3\right) \end{gathered}$$

Substitute all the known values in equation (2).

$$\begin{gathered} R=\frac{0.3\left(500\times {10}^{3}g\right)\left(6.022\times {10}^{23}mo{l}^{-1}\right)\left(24.9MeV\right)}{5\left(2.0g/mol\right)\left(2.6\times {10}^{28}MeV/megatonTNT\right)} \\ =\frac{2.249217\times {10}^{30}}{5\left(2.0g/mol\right)\left(2.6\times {10}^{18}MeV/megatonTNT\right)} \\ =8.65megatonTNT \end{gathered}$$

Therefore, the rating of the fusion part of the bomb is $8.65megatonTNT$.