How long can an electric lamp of 100; W be ke | vedclass

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(B) The given fusion reaction is: $_{1}^{2} H + _{1}^{2} H \rightarrow _{2}^{3} He + n + 3.27\; MeV$.Mass of deuterium,$m = 2.0\; kg = 2000\; g$.Numbe

Vedclass · Accepted Answer

$4.9 	imes 10^{4}\; years$

Vedclass · Answer

(B) The given fusion reaction is: $_{1}^{2} H + _{1}^{2} H ightarrow _{2}^{3} He + n + 3.27\; MeV$.Mass of deuterium,$m = 2.0\; kg = 2000\; g$.Number of moles in $2000\; g$ of deuterium = $\frac{2000}{2} = 1000\; moles$.Number of atoms in $2000\; g$ of deuterium = $1000 	imes 6.023 	imes 10^{23} = 6.023 	imes 10^{26}\; atoms$.Since $2\; atoms$ of deuterium release $3.27\; MeV$ of energy,the total energy $E$ released by $6.023 	imes 10^{26}\; atoms$ is:$E = \frac{6.023 	imes 10^{26}}{2} 	imes 3.27\; MeV = 3.0115 	imes 10^{26} 	imes 3.27\; MeV \approx 9.8476 	imes 10^{26}\; MeV$.Converting energy to Joules:$E = 9.8476 	imes 10^{26} 	imes 1.6 	imes 10^{-13}\; J = 1.5756 	imes 10^{14}\; J$.Power of the lamp $P = 100\; W = 100\; J/s$.Time $t = \frac{E}{P} = \frac{1.5756 	imes 10^{14}}{100} = 1.5756 	imes 10^{12}\; seconds$.Converting time to years:$t = \frac{1.5756 	imes 10^{12}}{60 	imes 60 	imes 24 	imes 365} \approx 4.99 	imes 10^{4}\; years \approx 4.9 	imes 10^{4}\; years$.

How long can an electric lamp of $100\; W$ be kept glowing by the fusion of $2.0\; kg$ of deuterium? Take the fusion reaction as $_{1}^{2} H + _{1}^{2} H \rightarrow _{2}^{3} He + n + 3.27\; MeV$.

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