Consider a beta decay reaction:{}_1^3H to {}_ | vedclass

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(C) The $Q$-value of the $\beta$ decay reaction is given by the mass difference between the parent nucleus and the products.Since the atomic mass of $

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$0.0186$

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(C) The $Q$-value of the $\beta$ decay reaction is given by the mass difference between the parent nucleus and the products.Since the atomic mass of ${}_1^3H$ includes one electron and the atomic mass of ${}_2^3He$ includes two electrons,we must account for the electron emitted in the decay.The $Q$-value is calculated as:$Q = [M({}_1^3H) - M({}_2^3He)] 	imes c^2$Note: The mass of the emitted electron is already accounted for in the atomic mass difference of the neutral atoms.$Q = (3.016050 \, u - 3.016030 \, u) 	imes 931.5 \, MeV/u$$Q = 0.000020 \, u 	imes 931.5 \, MeV/u$$Q = 0.01863 \, MeV$The maximum kinetic energy of the electron is approximately equal to the $Q$-value of the reaction.Therefore,the maximum energy is $0.0186 \, MeV$.

Consider a $\beta$ decay reaction:
${}_1^3H \to {}_2^3He + {e^{ - 1}} + \bar v$
The atomic masses of ${}_1^3H$ and ${}_2^3He$ are $3.016050 \, u$ and $3.016030 \, u$,respectively. Find the maximum possible energy of the electron in $MeV$.

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Consider a $\beta$ decay reaction:${}_1^3H \to {}_2^3He + {e^{ - 1}} + \bar v$The atomic masses of ${}_1^3H$ and ${}_2^3He$ are $3.016050 \, u$ and $3.016030 \, u$,respectively. Find the maximum possible energy of the electron in $MeV$.