Nuclei

Question: In the following reaction the value of ‘X’ is $_7{N^{14}}{ + _2}H{e^4}\, \to \,X{ + _1}{H^1}$

1) $_8{N^{17}}$

2) $_8{O^{17}}$

3) $_7{O^{16}}$

4) $_7{N^{16}}$

Solution:

Explanation: No Explanation

Nucleus Nuclear Reaction Modern Physics

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