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A deuteron is bombarded on $_{8}O^{16}$ nucleus then $\alpha$-particle is emitted then product nucleus is -
A
$_{7}N^{13}$
B
$_{5}B^{10}$
C
$_{4}Be^{9}$
D
$_{7}N^{14}$
Detailed Solution
Reaction: ${}^{16}_{8}O + {}^{2}_{1}H \rightarrow {}^{A}_{Z}X + {}^{4}_{2}He$
Use conservation of charge (atomic number) and mass number.
Mass number: $16 + 2 = A + 4 \Rightarrow A = 14$
Atomic number: $8 + 1 = Z + 2 \Rightarrow Z = 7$
Z = 7 is nitrogen, so the product nucleus is $_{7}N^{14}$.
Use conservation of charge (atomic number) and mass number.
Mass number: $16 + 2 = A + 4 \Rightarrow A = 14$
Atomic number: $8 + 1 = Z + 2 \Rightarrow Z = 7$
Z = 7 is nitrogen, so the product nucleus is $_{7}N^{14}$.