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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:.

For more information, see **Fission** | **Physics**

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