Correct Answer - 4.37 MeV
`""_(10)^(23)"Ne"to""_(11)^(23)"Na"+e^(-)+bar(v)+Q`.
`DeltaM=m(""_(10)^(20)"Ne")-[m(""_(11)^(23)"Na")+m(e^(-))]`
`=0.004696" u",m(e^(-))` का मान नगण्य होता है |
`:." "E_("max")=Delta" M.c"^(2)`
`=0.004696xx931.5("MeV")/(c^(2))xx c_(2)=4.37"MeV."`