A) \[_{\text{50}}^{\text{115}}\text{Sn}\]
B) \[_{\text{50}}^{\text{113}}\text{Sn}\]
C) \[_{49}^{\text{114}}\text{ln}\]
D) \[_{46}^{\text{115}}\text{Pa}\]
Correct Answer: A
Solution :
As we know that when \[\beta \]-particle is emitted the atomic number is increased by 1 unit. Hence by observing this fact by two successive \[\beta \]-decay we get\[{}_{50}S{{n}^{115}}\].
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