CUET Preparation Today
In the following reaction the value of ‘X’ is ${ }_7 N^{14}+{ }_2 H e^4 \rightarrow X+{ }_1 H^1$ |
${ }_8 N^{17}$ ${ }_8 O^{17}$ ${ }_7 O^{16}$ ${ }_7 N^{16}$ |
${ }_8 O^{17}$ |
${ }_7 N^{14}+{ }_2 He^4 \rightarrow{ }_8 O^{17}+{ }_1 H^1$ |
