Question

It is proposed to use the nuclear fusion reaction ${}_1{\text{H}}^2+{}_1{\text{H}}^2\to {}_2{\text{He}}^4$ in a nuclear reactor of $200\text{MW}$ rating. If $25\text{\%}$ of energy from the above reaction is used in the reactor. It is found that nearly $12x$ grams of deuterium will be needed per day. The value of $x$ is (integer only)

$[\text{Take,}\text{mass of}{}_1{\text{H}}^2=2.0141\text{amu},\text{mass of}{}_2{\text{He}}^4=4.0026\text{amu},1\text{amu}{c}^2=931.5\text{MeV}]$

Answer 1

Energy released in the nuclear fusion reaction is given by,

$Q=\left(\Delta m\right)c^2$

Here, $\Delta m=2M_{\left({}_1{\text{H}}^2\right)}-M_{\left({}_2{\text{He}}^4\right)}$

$\Rightarrow \Delta m=(2\times 2.0141u)-4.00264u=0.0256u=0.0256u\times c^2$

$\Rightarrow Q=0.0256\times 931.5\text{MeV}$

$\Rightarrow Q=23.8464\text{MeV}=23.8464\times 1.6\times {10}^{-13}\text{J}$

$\Rightarrow Q=38.15424\times {10}^{-13}\text{J}$

Since, only $25\text{\%}$ of energy is used in the reactor,

The effective energy used,

$Q^{\prime }=\frac{25}{100}\times Q$

$\Rightarrow Q^{\prime }=\frac{25}{100}\times 38.15424\times {10}^{-13}$

$\Rightarrow Q^{\prime }=9.53856\times {10}^{-13}\text{J}$

Two deuterium nuclei are involved in a fusion reaction,

$\therefore$ The energy released per deuterium

$E_1=\frac{9.53856\times {10}^{-13}}{2}\text{J}=4.76928\times {10}^{-13}\text{J}$

For $200\text{MW}$ power per day, the number of deuterium nuclear required,

$n=\frac{P\times t}{E_1}=\frac{200\times {10}^6\times 86400}{4.76928\times {10}^{-13}}[\because 1\text{day}=86400\text{sec}]$

$\Rightarrow n=3.62319\times {10}^{25}$

We know that, $6\times {10}^{23}$ atoms are present in $2\text{g}$ of deuterium.

Then, the mass of deuterium that contains $3.623\times {10}^{25}$ atoms is given by,

$m=\frac{2\times 3.62319\times {10}^{25}}{6\times {10}^{23}}\text{g}=120.77\text{g}\text{(or)}$

$m\approx 120\text{g}$

$\therefore x=10$