Practicing Success
Practicing Success
CUET Preparation Today
In the nuclear reaction: $X(n, \alpha)_3 L i^7$ the term X will be 3 |
${ }_5 B^{10}$ ${ }_5 B^9$ ${ }_5 B^{11}$ ${ }_2 He^4$ |
${ }_5 B^{10}$ |
$X(n, \alpha){ }_3^7 L i \Rightarrow{ }_Z X^A+{ }_0 n^1 \rightarrow 3 L i^7+{ }_2 He^4$ Z = 3 + 2 = 5 and A = 7 + 4 – 1 = 10 ∴ ${ }_5 X^{10} = { }_5 B^{10}$ |
