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A deuteron is bombard on ${ }_8 \mathrm{O}^{16}$ nuclei than $\alpha$-particle is emitted then the product nuclei is:
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Verified Answer
The correct answer is:
${ }_7 \mathrm{~N}^{14}$
${ }_8 \mathrm{O}^{16}+{ }_1 \mathrm{H}^1 \longrightarrow{ }_2 \mathrm{X}^4+{ }_2 \mathrm{He}^4$
Use conservation of charge and mass
The conservation of mass number gives
$$
16+2=A+4 \Rightarrow A=14
$$
The conservation of atomic number gives.
$$
8+1=2+2 \Rightarrow 2=7
$$
Therefore, ${ }_2 \mathrm{X}^4$ is ${ }_7 \mathrm{~N}^{14}$
Use conservation of charge and mass
The conservation of mass number gives
$$
16+2=A+4 \Rightarrow A=14
$$
The conservation of atomic number gives.
$$
8+1=2+2 \Rightarrow 2=7
$$
Therefore, ${ }_2 \mathrm{X}^4$ is ${ }_7 \mathrm{~N}^{14}$
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