The isotope {}_{5}^{12}B having a mass 12.014 | vedclass

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Question

(B) The $\beta$-decay process is given by: ${}_{5}^{12}B 	o {}_{6}^{12}C^* + e^- + \bar{
u}_e$.The $Q$-value of the decay to the ground state of ${}

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The $\beta$-decay process is given by: ${}_{5}^{12}B 	o {}_{6}^{12}C^* + e^- + \bar{
u}_e$.The $Q$-value of the decay to the ground state of ${}_{6}^{12}C$ is calculated using the mass difference: $Q = [M({}_{5}^{12}B) - M({}_{6}^{12}C)] 	imes 931.5 	ext{ MeV/u}$.Assuming the mass of ${}_{6}^{12}C$ is approximately $12.000 	ext{ u}$ (standard carbon-$12$ mass), the total energy available is $Q = (12.014 - 12.000) 	imes 931.5 	ext{ MeV} = 0.014 	imes 931.5 	ext{ MeV} \approx 13.041 	ext{ MeV}$.Since the decay is to the excited state ${}_{6}^{12}C^*$, which is $4.041 	ext{ MeV}$ above the ground state, the energy available for the $\beta$-particle and the antineutrino is $Q' = Q - 4.041 	ext{ MeV} = 13.041 - 4.041 = 9 	ext{ MeV}$.The maximum kinetic energy of the $\beta$-particle occurs when the antineutrino energy is zero, which is equal to $Q'$.Therefore, the maximum kinetic energy is $9 	ext{ MeV}$.

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