A) \[_{-1}^{0}e\]
B) \[_{1}^{1}H\]
C) \[_{1}^{2}H\]
D) \[_{0}^{1}n\]
Correct Answer: D
Solution :
In the reaction, we have to conserve the mass number and the atomic number. Thus, we have \[_{7}^{14}N+_{b}^{a}X\xrightarrow{{}}_{6}^{14}C+_{1}^{1}H\] So, \[14+a=14+1\] Hence a = 1 and 7 + b = 6 + 1 Hence b = 0. So, we have \[_{b}^{a}X=_{0}^{1}X=_{0}^{1}n.\]You need to login to perform this action.
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