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Olympiad workbook
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Olympiad workbook
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NSEP
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Nuclear Physics

Olympiad workbook · NSEP · Nuclear Physics

The Nucleus _ 10 ^ 23 Ne decays by β- emission through the reaction...

The Nucleus 1023Ne{ }_{10}^{23} \mathrm{Ne} decays by β−\beta- emission through the reaction 1023Ne→1123Na+−10β+vˉ+{ }_{10}^{23} \mathrm{Ne} \rightarrow{ }_{11}^{23} \mathrm{Na}+{ }_{-1}^0 \beta+\bar{v}+ energy The atomic masses are 1023Ne=22.994466u{ }_{10}^{23} \mathrm{Ne}=22.994466 \mathrm{u} and 1123Na=22.989770u,−10β=0.000549u{ }_{11}^{23} N a=22.989770 \mathrm{u},{ }_{-1}^0 \beta=0.000549 \mathrm{u}. The maximum kinetic energy that the emitted electron can ever have is

Options

  1. A4.374 MeV
  2. B3.862 MeV
  3. C2.187 MeV
  4. D1.931 MeV

Correct answer

A. 4.374 MeV

Step-by-step solution

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