The nucleus _{48}^{115}Cd after two successiv | vedclass

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(D) In a ${\beta ^ - }$ decay,the atomic number $Z$ increases by $1$ while the mass number $A$ remains constant. The process is represented as: $_{Z}^

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$_{50}^{115}Sn$

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(D) In a ${\beta ^ - }$ decay,the atomic number $Z$ increases by $1$ while the mass number $A$ remains constant. The process is represented as: $_{Z}^{A}X ightarrow _{Z+1}^{A}Y + _{-1}^{0}e + \bar{
u}$.Starting with $_{48}^{115}Cd$:First ${\beta ^ - }$ decay: $_{48}^{115}Cd ightarrow _{49}^{115}In + _{-1}^{0}e + \bar{
u}$.Second ${\beta ^ - }$ decay: $_{49}^{115}In ightarrow _{50}^{115}Sn + _{-1}^{0}e + \bar{
u}$.Thus,after two successive ${\beta ^ - }$ decays,the nucleus becomes $_{50}^{115}Sn$.

The nucleus $_{48}^{115}Cd$ after two successive ${\beta ^ - }$ decays will give:

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