$\ _{7}N^{14}\ $is bombarded with $\ _{2}He^{4}.$ The resulting nucleus is $\ _{8}O^{17}$ with the emission of

Question

Dr. Abubacker · Accepted Answer

(c) The nuclear reactions is as follows
$\ _{7}N^{14}\  + \ \ _{2}He^{4}\  \rightarrow \ _{8}O^{17}\  + \ \ _{q}X^{p}$
Conservation of mass number gives
$\ \ \ \ \ \ \ \ \ \ P = 14 + 4 - 17 = 1$
Conservation of atomic number gives
$\ \ \ \ \ \ \ \ \ \ \ a = 7 + 2 - 8 = 1$
Hence, particle is a proton $\ _{1}H^{1}.$

Q. $\ _{7}N^{14}\ $is bombarded with $\ _{2}He^{4}.$ The resulting nucleus is $\ _{8}O^{17}$ with the emission of

Explanation

Reviewed by Dr. Abubacker

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