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  4. N 14 is bombarded with 2 He 4. The resulting nucleus is 8 O 17 with the emission of

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

Bihar CECEBihar CECE 2007Atoms

Solution:

Mass number and atomic number in nuclear reactions are conserved.
The nuclear reaction is as follows.
${ }_{7} N ^{14}+{ }_{2} He ^{4} \longrightarrow { }_{8} O ^{17}+{ }_{q} X^{P}$
Conservation of mass number gives
$p=14+4-17=1$
Conservation of atomic number gives
$q=7+2-8=1$
Hence, particle is a proton $_{1} H ^{1}$.