Question

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

• A. ${}_{46}^{115}Pa$
• B. ${}_{49}^{114}In$
• C. ${}_{50}^{113}Sn$
• D. ${}_{50}^{115}Sn$

Answer 1

Answer: (D)

${}_{50}^{115}Sn$

Answer 2

$48Cd^{115}{\overset{\quad 2( - 1\beta^{o})\quad}{\rightarrow}}_{50}Sn^{115}$