You are given that $$^7_3\text{Li} = 7.0160\,\text{u}$$, Mass of $$^4_2\text{He} = 4.0026\,\text{u}$$ and Mass of $$^1_1\text{He} = 1.0079\,\text{H}$$. When $$20\,\text{g}$$ of $$^7_3\text{Li}$$ is converted into $$^4_2\text{He}$$ by proton capture, the energy liberated, (in kWh), is: [Mass of nucleon $$= 1\,\text{GeV}/c^2$$]

We are told that the nuclear reaction is a proton-capture process in which a lithium-7 nucleus absorbs one proton and breaks up into two alpha particles. Symbolically we write

$$\,^{7}_{3}\text{Li}+\,^{1}_{1}\text{H}\;\longrightarrow\;2\,^{4}_{2}\text{He}+Q.$$

Here $$Q$$ is the energy released. According to Einstein’s mass-energy equivalence,

$$Q=\Delta m\,c^{2},$$

where the mass defect $$\Delta m$$ is the difference between the total mass of the initial nuclei and the total mass of the final nuclei.

We now calculate every quantity step by step.

The given atomic masses are

$$m\!\left(\,^{7}_{3}\text{Li}\right)=7.0160\;\text{u},\quad m\!\left(\,^{1}_{1}\text{H}\right)=1.0079\;\text{u},\quad m\!\left(\,^{4}_{2}\text{He}\right)=4.0026\;\text{u}.$$

First we obtain the total initial mass:

$$m_{\text{initial}} =m\!\left(\,^{7}_{3}\text{Li}\right)+m\!\left(\,^{1}_{1}\text{H}\right) =7.0160\;\text{u}+1.0079\;\text{u} =8.0239\;\text{u}.$$

Next we calculate the total final mass. Two alpha particles are produced, so

$$m_{\text{final}} =2\,m\!\left(\,^{4}_{2}\text{He}\right) =2\times4.0026\;\text{u} =8.0052\;\text{u}.$$

The mass defect is therefore

$$\Delta m =m_{\text{initial}}-m_{\text{final}} =8.0239\;\text{u}-8.0052\;\text{u} =0.0187\;\text{u}.$$

We convert this mass defect into energy. The numerical relation used in nuclear physics is

$$1\;\text{u}=931.5\;\text{MeV}/c^{2}.$$

Hence, energy released per reaction is

$$Q =\Delta m\,c^{2} =0.0187\;\text{u}\times931.5\;\frac{\text{MeV}}{\text{u}} =17.4\;\text{MeV}\;(\text{approximately}).$$

To proceed toward macroscopic energy we translate this value into joules. Using

$$1\;\text{eV}=1.602\times10^{-19}\;\text{J},$$

we have

$$Q =17.4\times10^{6}\;\text{eV}\times1.602\times10^{-19}\;\frac{\text{J}}{\text{eV}} =2.79\times10^{-12}\;\text{J}.$$

Now we find how many such reactions occur when $$20\;\text{g}$$ of lithium-7 is completely converted.

The molar mass of $$\,^{7}_{3}\text{Li}$$ is $$7\;\text{g mol}^{-1}$$, so the number of moles in $$20\;\text{g}$$ is

$$n =\frac{20\;\text{g}}{7\;\text{g mol}^{-1}} =2.857\;\text{mol}.$$

Using Avogadro’s number $$N_{\!A}=6.022\times10^{23}\;\text{mol}^{-1}$$, the number of lithium nuclei is

$$N =n\,N_{\!A} =2.857\;\text{mol}\times6.022\times10^{23}\;\frac{\text{nuclei}}{\text{mol}} =1.72\times10^{24}\;\text{nuclei}.$$

Each nucleus undergoes one reaction, so the total energy liberated is

$$E_{\text{total}} =N\,Q =1.72\times10^{24}\times2.79\times10^{-12}\;\text{J} =4.80\times10^{12}\;\text{J}.$$

The problem asks for the answer in kilowatt-hours (kWh). We recall the conversion

$$1\;\text{kWh}=3.6\times10^{6}\;\text{J}.$$

Therefore,

$$E_{\text{total}}\;(\text{kWh}) =\frac{4.80\times10^{12}\;\text{J}}{3.6\times10^{6}\;\text{J kWh}^{-1}} =1.33\times10^{6}\;\text{kWh}.$$

Hence, the correct answer is Option D.