The radioactive nuclide decays according to 116C→115B+e++ν.The disi – InterviewSolution

1.	The radioactive nuclide decays according to 116C→115B+e++ν.The disintegration energy of this process is Q(in MeV). Given that atomic masses mC=11.011433u,me=0.0005486u,mB=11.009305u,1 amu=931 MeV. Then the value of 100Q19 is
Answer» The radioactive nuclide decays according to $_{6}^{11} C \to_{5}^{11} B + e^{+} + ν$ .The disintegration energy of this process is $Q$ (in $M e V$ ). Given that atomic masses $m_{C} = 11.011433 u, m_{e} = 0.0005486 u, m_{B} = 11.009305 u, 1 a m u = 931 M e V$ . Then the value of $\frac{100 Q}{19}$ is

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