Nuclear reactions are given as:(i) , square ( | vedclass

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(A) To find the missing particle or nuclide,we apply the conservation of mass number $(A)$ and atomic number $(Z)$.$(i) \, {_{Z}}X^{A} + {_{0}}n^{1} \

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$S^{32}, F^{19}, {_{0}}n^{1}$

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(A) To find the missing particle or nuclide,we apply the conservation of mass number $(A)$ and atomic number $(Z)$.$(i) \, {_{Z}}X^{A} + {_{0}}n^{1} \rightarrow {_{15}}P^{32} + {_{1}}H^{1}$Equating $Z$: $Z + 0 = 15 + 1 \Rightarrow Z = 16$.Equating $A$: $A + 1 = 32 + 1 \Rightarrow A = 32$.Thus,the nuclide is ${_{16}}S^{32}$.$(ii) \, {_{Z}}X^{A} + {_{1}}H^{1} \rightarrow {_{8}}O^{16} + {_{2}}He^{4}$Equating $Z$: $Z + 1 = 8 + 2 \Rightarrow Z = 9$.Equating $A$: $A + 1 = 16 + 4 \Rightarrow A = 19$.Thus,the nuclide is ${_{9}}F^{19}$.$(iii) \, {_{7}}N^{14} + {_{Z}}X^{A} \rightarrow {_{6}}C^{14} + {_{1}}H^{1}$Equating $Z$: $7 + Z = 6 + 1 \Rightarrow Z = 0$.Equating $A$: $14 + A = 14 + 1 \Rightarrow A = 1$.Thus,the particle is ${_{0}}n^{1}$ (neutron).Therefore,the missing values are ${_{16}}S^{32}, {_{9}}F^{19}, {_{0}}n^{1}$.

Nuclear reactions are given as:
$(i) \, \square (n, p) {_{15}}P^{32}$
$(ii) \, \square (p, \alpha) {_{8}}O^{16}$
$(iii) \, {_{7}}N^{14} (\square, p) {_{6}}C^{14}$
The missing particles or nuclides (in box $\square$) in these reactions are respectively:

Similar Questions

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Nuclear reactions are given as:$(i) \, \square (n, p) {_{15}}P^{32}$$(ii) \, \square (p, \alpha) {_{8}}O^{16}$$(iii) \, {_{7}}N^{14} (\square, p) {_{6}}C^{14}$The missing particles or nuclides (in box $\square$) in these reactions are respectively: