The radionuclide ^{11}_{6}C decays by beta^+ | vedclass

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Question

(A) The $\beta^+$-decay equation for $^{11}_{6}C$ is:$^{11}_{6}C \longrightarrow ^{11}_{5}B + ^{0}_{+1}e + 
u + Q$To calculate the $Q$-value,we use t

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$0.962 \ MeV$

Vedclass · Answer

(A) The $\beta^+$-decay equation for $^{11}_{6}C$ is:$^{11}_{6}C \longrightarrow ^{11}_{5}B + ^{0}_{+1}e + 
u + Q$To calculate the $Q$-value,we use the atomic masses. The decay of a nucleus $^{A}_{Z}X$ into $^{A}_{Z-1}Y$ via $\beta^+$ emission is given by:$Q = [m(^{A}_{Z}X) - m(^{A}_{Z-1}Y) - 2m_e]c^2$Substituting the given values:$Q = [11.011434 \ u - 11.009305 \ u - 2(0.000548 \ u)] 	imes 931.5 \ MeV/u$$Q = [11.011434 - 11.009305 - 0.001096] 	imes 931.5 \ MeV$$Q = [0.001033] 	imes 931.5 \ MeV$$Q \approx 0.962 \ MeV$

The radionuclide $^{11}_{6}C$ decays by $\beta^+$ emission. Given that $m(^{11}_{6}C) = 11.011434 \ u$,$m(^{11}_{5}B) = 11.009305 \ u$,$m_e = 0.000548 \ u$,and $1 \ u = 931.5 \ MeV/c^2$. The $Q$-value of this decay process is:

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