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 \[{}_{1}{{H}^{2}}+{}_{1}{{H}^{2}}\xrightarrow{{}}{}_{2}H{{e}^{3}}+n+3.2\,MeV\]

Answer:

\[t=4.9\times {{10}^{4}}\] 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                         \[=\frac{\text{Energy}}{\text{Power}}\]                         \[=\frac{15.42\times {{10}^{13}}}{100}\]                         \[=15.42\times {{10}^{11}}\,\,\text{seconds}\]                         \[=4.9\times {{10}^{4}}\,\,\text{years}\]