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The radionuclide $$^{56}Mn$$ has a half-life of 2.58 h and is produced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manganese isotope $$^{55}Mn$$, and the manganese–deuteron reaction that produces $$ ^{56}Mn$$ is
                $$^{55}Mn + d \rightarrow ^{56}Mn +p $$
 If the bombardment lasts much longer than the half-life of $$^{56}Mn$$, the activity of the $$^{56}Mn$$ produced in the target reaches a final value of $$8.88 \times 10^10 Bq$$. (a) At what rate is $$^{56}Mn$$ being produced? (b) How many $$^{56}$$Mn nuclei are then in the target? (c) What is their total mass?


Solution

 If $$M_He$$ is the mass of an atom of helium and $$M_C$$ is the mass of an atom of carbon, then 
the energy released in a single fusion event is
               $$ Q= (3M_He - Mc)c^2 = 3 [3(4.0026 u) (12.0000 u)](931.5 MeV/u)= 7.27 MeV$$.
 Note that $$3M_He$$ contains the mass of six electrons and so does $$M_C$$. The electron masses 
cancel and the mass difference calculated is the same as the mass difference of the nuclei.

Physics
NCERT
Standard XII

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