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Consider the following nuclear fission ...

Consider the following nuclear fission reaciton
`._(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

Mass defect `Deltam=226.005 -222.000-4.000`
=0.005 amu
`:. ` Q value `=0.005xx431.5 =4.655 MeV`
Also `(K.E_(alpha))/(K.E_(Rn))=(m_(Rn))/(m_(alpha))`
`implies K.E_(Rn)=(m_(alpha))/(m_(Rn)). K.E_(alpha) =4/222xx4.44 =0.08 MeV`
`:. ` Energy of `gamma`-photon =4.655-(4.44+0.08)
=0.135 MeV
=135 KeV
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