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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Solution

For the given reaction mass defect is
$Δ m = [m_{_{6}^{11} C} - 6 m_{e}] - [m_{_{5}^{11} B} - 5 m_{e} + m_{e}]$
$Δ m = [m_{_{6}^{11} C} - m_{_{5}^{11} B} - 2 m_{e}]$
$Δ m = 0.0010308 u$
For positron emission
$Q = (Δ m) 931 M e V$
$Q = (0.0010308) 931 = 0.9596748 MeV$
Hence $\frac{100 Q}{19} = 5.05$

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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

For the given reaction mass defect is Δm=[m116C−6me]−[m115B−5me+me] Δm=[m116C−m115B−2me] Δm=0.0010308u For positron emission Q=(Δm)931 MeV Q=(0.0010308)931=0.9596748 MeV Hence 100Q19=5.05