• # question_answer How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction can be taken as ${}_{1}{{H}^{2}}+{}_{1}{{H}^{2}}\xrightarrow{{}}{}_{2}H{{e}^{3}}+n+3.2\,MeV$

$t=4.9\times {{10}^{4}}$ years According to the concept of Avogadro number, the number of atoms in 2g of $_{1}^{2}H=6.023\times {{10}^{23}}$ $\therefore$ The number of atoms in 2 kg of $_{1}^{2}H$             $=\frac{6.023\times {{10}^{23}}}{2}\times 2000$ $=6.023\times {{10}^{26}}\,\,\text{atoms}$ The energy released in fusion of 2 atoms = 32 MeV $\therefore$ Total energy released in fusion of 2 kg of deuterium             $=\frac{6.023\times {{10}^{23}}}{2}\times 3.2\,\,\text{MeV}$ $=3.0115\times {{10}^{26}}\times 32\times 1.6\times {{10}^{-19}}\times {{10}^{6}}\,\,\text{J}$ $=15.42\times {{10}^{13}}\,\,\text{J}$             Power rating of the electric lamp                         $=100\,\,\text{W}$             Energy = Power $\times$ time             $\therefore$ Time for which the lamp glows                         $=\frac{\text{Energy}}{\text{Power}}$                         $=\frac{15.42\times {{10}^{13}}}{100}$                         $=15.42\times {{10}^{11}}\,\,\text{seconds}$                         $=4.9\times {{10}^{4}}\,\,\text{years}$

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