Question

Verify that the fusion of1.0 kgof deuterium by the reaction${\mathbf{}}^{\mathbf{2}}\mathbf{H}\mathbf{+}{\mathbf{}}^{\mathbf{2}}\mathbf{H}\mathbf{\to }{\mathbf{}}^{\mathbf{3}}\mathbf{He}\mathbf{+}\mathbf{n}\mathbf{(}\mathbf{Q}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{27}\mathbf{MeV}\mathbf{)}$could keep a100 Wlamp burning for$\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{y}$.

Short answer

It is verified that the fusion of 1.0 kg deuterium by the reaction could keep a 100 W lamp burning for $2.5\times {10}^4\mathrm{y}$.

Step-by-step solution

The given data

1. Mass of deuterium, m = 1.0 kg or 1000g
2. The given reaction, ${}^2\mathrm{H}+{}^2\mathrm{H}\to {}^3\mathrm{He}+\mathrm{n}$
3. The energy released by the reaction or the Q-value, Q = +3.27 MeV
4. Power of the lamp, P = 100W

Understanding the concept of fusion reaction

In a nuclear fusion reaction, two or more nuclei combine to produce a heavy nucleus with some energy being released. This energy is given by the Q-value of the reaction. The Q-value of the fusion process is the amount of energy that is either absorbed or released during the process. Thus, it describes the integration process for a fusion process that involves the binding of two nuclei.

Formulae:

The power generated due to the energy released by the reactor is as folows:

$P=\frac{E}{t}$ ….. (i)

Write the number of particles in an atomas follows:

$N=\frac{m}{M}{N}_A$ …… (ii)

$\mathrm{Here},N_A=6.022\times {10}^{23}{\mathrm{mol}}^{-1}$

Here, m is the given mass and M is the molar mass of the atom.

Verify the time of the disintegration process of the given reaction

Given the energy release per fusion$(\mathrm{Q}=3.27\mathrm{MeV}\mathrm{or}5.24\times {10}^{-13}\mathrm{J})$and that a pair of deuterium atoms is consumed in each fusion event.
Now, the pairs of deuterium atoms consumed in the given reaction is given using equation (ii) as follows:

$$\begin{gathered} \mathrm{N}=\frac{1000}{2(2.0\mathrm{g}.\mathrm{mol})}(6.022\times {10}^{23}/\mathrm{mol})(\because \mathrm{We}\mathrm{are}\mathrm{calculating}\mathrm{the}\mathrm{deuterium}\mathrm{pairs}) \\ =1.5\times {10}^{26} \end{gathered}$$

Now, using the above number of deuterium pairs, the total energy released is given as:

$$\begin{gathered} E_{total}=\left(1.5\times {10}^{26}\right)\left(5.24\times {10}^{-13}\mathrm{J}\right) \\ =7.9\times {10}^{13}\mathrm{J} \end{gathered}$$

Thus, using this value in equation (i), the required time of the decay process or the time of the fusion process is calculated as follows:

$$\begin{gathered} \mathrm{t}=\frac{7.9\times {10}^{13}\mathrm{J}}{100\mathrm{W}} \\ =7.9\times {10}^{11}\mathrm{s} \\ =2.5\times {10}^4\mathrm{y} \end{gathered}$$

Hence, it is verified that the value of the required time is $2.5\times {10}^4\mathrm{y}$.