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(B) In $\beta^{-}$ decay, the process is represented as:${ }_{Z}^{A} X \longrightarrow { }_{Z+1}^{A} Y + { }_{-1}^{0} \beta + \bar{
u}$If only an ele

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$(a)$ in case $I$ and $(b)$ in case $II$

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(B) In $\beta^{-}$ decay, the process is represented as:${ }_{Z}^{A} X \longrightarrow { }_{Z+1}^{A} Y + { }_{-1}^{0} \beta + \bar{
u}$If only an electron were emitted (Case $I$), the electron would carry a fixed amount of energy, resulting in a sharp, monoenergetic peak as shown in curve $(a)$.However, in reality (Case $II$), the decay energy is shared between the emitted electron $(\beta^{-})$ and the antineutrino $(\bar{
u})$. Because the energy is distributed between two particles, the emitted electrons have a continuous range of energies, resulting in the broad spectrum shown in curve $(b)$.Therefore, Case $I$ corresponds to $(a)$ and Case $II$ corresponds to $(b)$.

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