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Consider the following nuclear fission ...

Consider the following nuclear fission reaction
`._(88)Ra^(226) to ._(86)Rn^(222)+._(2)He^(4)+Q`
In the fission reaction. Kinetic energy of `alpha`-particle is 4.44 MeV. Find the energy emitted as `gamma`-radiation in keV in this reaction.
`m(._(88)Ra^(226))=226.005` amu
`m(._(86)Rn^(222))=222.000` amu

Text Solution

Verified by Experts

`Q=[M(Ra^(226)-M(Rn^(222))-M(He^(4))]xx931`
`=(226.025406-222.017574-4.002603)uxx931`
`=0.005229 uxx931(Mev)/u`
Q=4.87 MeV
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