How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as ^{2}_1H + ^{2}_{1} H \rightarrow ^3_2 He + n + 3.27 MeV

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Fission | Physics
Hundreds of nuclear fission power plants around the world attest to the fact that ... Fission is the opposite of fusion and releases energy only when heavy nuclei are split. ... The amount of energy per fission reaction can be large, even by nuclear ... Calculate the energy released in the following spontaneous fission reaction:.
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Given that
Power P = 100 W
Mass of the deuterium = 2.0 kg
Nuclear reaction: ^{2}_1 H + ^{2}_1 H \rightarrow ^{3}_{2} He + n + 3.27 MeV
Step 1: Finding the number of atoms in the 2 kg fission matter
Avogadro number NA = 6.023 * 10^{23}
Atomic mass of deuterium =\ 2
Mass of a deuterium m=2\ kg\ =\ 2000\ g
Number of atoms in the 2kg deuterium,\ N=\frac{NA}{A}*m
= \frac{6.023*10^{23}}{2}* 2 kg
= 6.023*10^{26} atoms
Step 2: Determining the total energy released during the fission of atoms in 2kg
For two deuterium under fusion release 3.27 MeV
Average energy per fission = \frac{3.27}{2} = 1.635
Total energy E = Average energy per fusion * Number of atoms
E = 1.635 * 6.023*10^{26}
E = 9.85* 10^{26} MeV
E = 9.85* 10^{26} * 1.6* 10^{-19} * 10^{6} Joules
E = 1.576 * 10^{14} Joules
Step 3: Detecting the length(time) of an electric lamp's glow
The amount of energy divided by the time equals power.
P = \frac{E}{t}
t = \frac{E}{P}
t = \frac{ 1.576 * 10^{14}}{100}
t = 1.576 * 10^{12} seconds
t = \frac{1.576 * 10^{12}}{365* 60*60* 24}
t = 4.9 * 10^{4} years
Hence, The amount of time it takes for the lamp to glow t = 4.9 * 10^{4} years