2147Sc→2247Ti+−10βand2147Sc→2047Ca+10β+{}_{21}^{47}\text{Sc} \rightarrow {}_{22}^{47}\text{Ti} + {}_{-1}^{0}\beta \quad \text{and} \quad {}_{21}^{47}\text{Sc} \rightarrow {}_{20}^{47}\text{Ca} + {}_{1}^{0}\beta^{+}…

Problem

2147Sc→2247Ti+−10βand2147Sc→2047Ca+10β+{}_{21}^{47}\text{Sc} \rightarrow {}_{22}^{47}\text{Ti} + {}_{-1}^{0}\beta \quad \text{and} \quad {}_{21}^{47}\text{Sc} \rightarrow {}_{20}^{47}\text{Ca} + {}_{1}^{0}\beta^{+}The following data are available:
ParticleMass in amu
2147Sc{}_{21}^{47}\text{Sc}46.9524
2047Ca{}_{20}^{47}\text{Ca}46.9545
2247Ti{}_{22}^{47}\text{Ti}46.9518
−10β{}_{-1}^{0}\beta and 10β+{}_{1}^{0}\beta^{+}0.00055
If the initial 2147Sc{}_{21}^{47}\text{Sc} nucleus is considered to be stationary, both of these equations suggest that the products of the reaction will gain kinetic energy and move away from each other.

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