By Sunil Bhardwaj

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In 1939; Ottohahn and Strassman, established that radio active isotope of barium, Lanthanum, Strontium, Yttrium, rubidium and cesiums, were formed as a result of uranium nucleus splitting into two (not necessarily equal) parts following neutrons capture.

Meitner and Frisch carried out detailed calculations and predicted that in the splitting process, which they called fission an energy of about 200 MeV (equivalent to 0.21 amu ) should be set free by each splitting uranium nucleus. This prediction was later experimentally confirmed by Frisch.

In 1939, Bhor and Wheeler provided the mechanism of the Fission reaction in terms of liquid drop model. They also established that it was rare isotopes of uranium U235, which underwent fission. Since then it has been established that fission can be induced in several other heavy nuclei through interaction of neutrons. In reality the fission fragments of U253 fission fall into two groups - a light group with atomic number ranging from 85 - 104 and heavy group with atomic number ranging from 130-149. The energy released during a fission can be calculated using exact mass difference between the reactants and products.

Let us consider specific case where Mo95 and La139 are two stable end products of fission. The reaction is $$ _{ 92 }^{ 235 }{ U }+_{ 0 }^{ 1 }{ n }\underset { slow }{ \longrightarrow } \left[ _{ 92 }^{ 236 }{ U } \right] \longrightarrow _{ 42 }^{ 95 }{ Mo }+_{ 57 }^{ 139 }{ La }+2_{ 0 }^{ 1 }{ n }+7{ e }^{ - }+Energy $$ For this reaction,

the total mass of the reactants = 236.0526 amu.

and the total mass of the products = 235.8329 amu.

Thus the mass difference = 0:2197 amu:

The energy released = 0.2197 \(\times\) 931 = 204.5 MeV.

Let us consider specific case where \({ Mo }^{ 95 }\) and \({ La }^{ 139 }\) are two stable end products of fission.

The reaction is, $$ _{ 92 }^{ 235 }{ U } + _{ 0 }^{ 1 }{ n } \underset { slow }{ \longrightarrow } \left[ _{ 92 }^{ 236 }{ U } \right] \longrightarrow _{ 42 }^{ 95 }{ Mo } + _{ 57 }^{ 139 }{ La } + 2_{ 0 }^{ 1 }{ n } + 7{ e }^{ - } + Q $$ Mass of \(_{ 92 }^{ 235 }{ U } = 235.0439 amu\)

Mass of \(_{ 0 }^{ 1 }{ n } = 1.0087 amu\)

\(\therefore \) The total mass of the reactants \(= 235.0439 + 1.0087 = 236.0526 amu.\)

Mass of \(_{ 42 }^{ 95 }{ Mo } = 94.9058 amu\)

Mass of \(_{ 57 }^{ 139 }{ La } = 138.9064 amu\)

Mass of \(_{ 0 }^{ 1 }{ n } = 1.0087 amu\)

Mass of \({ e }^{ - } = 0.0005 amu\)

\( \therefore \) The total mass of the products \(= 94.9058 + 138.9064 + \left( 2\times 1.0087 \right) + \left( 7\times 0.0005 \right) \)

\(= 235.8331amu.\)

Thus the mass difference \( = 236.0526 - 235.8331 = 0.2195 amu\)

The energy released \(= 0.2195 \times 931 = 204.4 MeV.\)