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The isotope (5)^(12) B having a mass 12....

The isotope `(5)^(12) B` having a mass `12.014 u`undergoes beta - decay to `_(6)^(12) C _(6)^(12) C `has an excited state of the nucleus `( _(6)^(12) C ^(**) at 4.041 MeV` above its ground state if `_(5)^(12)E` decay to `_(6)^(12) C ^(**) ` , the maximum kinetic energy of the `beta` - particle in unit of MeV is `(1 u = 931.5MeV//c^(2)` where c is the speed of light in vaccuum) .

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To solve the problem of finding the maximum kinetic energy of the beta particle emitted during the beta decay of the isotope \( _{5}^{12}B \) to the excited state of \( _{6}^{12}C^* \), we will follow these steps: ### Step 1: Identify the Masses We know the mass of the boron isotope: - Mass of \( _{5}^{12}B = 12.014 \, u \) - Mass of \( _{6}^{12}C \) (ground state) is approximately \( 12.000 \, u \). ### Step 2: Calculate the Mass Defect The mass defect \( \Delta m \) can be calculated as: \[ \Delta m = \text{Mass of reactant} - \text{Mass of products} \] In this case, the products include the mass of the carbon nucleus and the emitted beta particle (electron) and antineutrino. However, for simplicity, we can ignore the small mass of the emitted particles in this calculation. Thus, we have: \[ \Delta m = 12.014 \, u - 12.000 \, u = 0.014 \, u \] ### Step 3: Calculate the Q-value The Q-value of the reaction can be calculated using the mass defect: \[ Q = \Delta m \times 931.5 \, \text{MeV/c}^2 \] Substituting the value of \( \Delta m \): \[ Q = 0.014 \, u \times 931.5 \, \text{MeV/u} = 13.041 \, \text{MeV} \] ### Step 4: Account for the Excitation Energy The excited state of \( _{6}^{12}C \) has an excitation energy of \( 4.041 \, \text{MeV} \). Therefore, the kinetic energy \( K \) of the emitted beta particle can be found using the conservation of energy: \[ K = Q - \text{Excitation Energy} \] Substituting the known values: \[ K = 13.041 \, \text{MeV} - 4.041 \, \text{MeV} = 9.000 \, \text{MeV} \] ### Final Answer The maximum kinetic energy of the beta particle emitted during the decay is: \[ \boxed{9 \, \text{MeV}} \]

To solve the problem of finding the maximum kinetic energy of the beta particle emitted during the beta decay of the isotope \( _{5}^{12}B \) to the excited state of \( _{6}^{12}C^* \), we will follow these steps: ### Step 1: Identify the Masses We know the mass of the boron isotope: - Mass of \( _{5}^{12}B = 12.014 \, u \) - Mass of \( _{6}^{12}C \) (ground state) is approximately \( 12.000 \, u \). ### Step 2: Calculate the Mass Defect ...
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