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Question:
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Consider the following nuclear reactions :
I. $\quad{ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \rightarrow_{8}^{17} \mathrm{O}+\mathrm{X}$
II. ${ }_{4}^{9} \mathrm{Be}+{ }_{2}^{4} \mathrm{H} \rightarrow_{6}^{12} \mathrm{He}+\mathrm{Y}$
Then
Options:
I. $\quad{ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \rightarrow_{8}^{17} \mathrm{O}+\mathrm{X}$
II. ${ }_{4}^{9} \mathrm{Be}+{ }_{2}^{4} \mathrm{H} \rightarrow_{6}^{12} \mathrm{He}+\mathrm{Y}$
Then
Solution:
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Verified Answer
The correct answer is:
$\mathrm{X}$ is a proton and $\mathrm{Y}$ is a neutron.
${ }_{7} \mathrm{~N}^{14}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{8} \mathrm{O}^{17}+{ }_{1} \mathrm{H}^{1}$
${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{H}^{4} \rightarrow{ }_{6} \mathrm{He}^{12}+{ }_{0} \mathrm{n}^{1}$
${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{H}^{4} \rightarrow{ }_{6} \mathrm{He}^{12}+{ }_{0} \mathrm{n}^{1}$
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