Nuclear reactions are given as (i) (n, p) {}_ | vedclass

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(A) To find the missing particles,we apply the law of conservation of mass number and atomic number.$(i)$ The reaction is $(n, p)$ on a target to prod

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${}_{16}S^{32}, {}_{9}F^{19}, {}_{0}n^{1}$

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(A) To find the missing particles,we apply the law of conservation of mass number and atomic number.$(i)$ The reaction is $(n, p)$ on a target to produce ${}_{15}P^{32}$. The reaction is ${}_{16}S^{32} + {}_{0}n^{1} 	o {}_{15}P^{32} + {}_{1}H^{1}$. Thus,the target is ${}_{16}S^{32}$.$(ii)$ The reaction is $(p, \alpha)$ on a target to produce ${}_{8}O^{16}$. The reaction is ${}_{9}F^{19} + {}_{1}H^{1} 	o {}_{2}He^{4} + {}_{8}O^{16}$. Thus,the target is ${}_{9}F^{19}$.$(iii)$ The reaction is ${}_{7}N^{14} + ? 	o {}_{6}C^{14} + {}_{1}H^{1}$. Balancing mass numbers: $14 + A = 14 + 1 \implies A = 1$. Balancing atomic numbers: $7 + Z = 6 + 1 \implies Z = 0$. The missing particle is ${}_{0}n^{1}$ (neutron).Therefore,the missing components are ${}_{16}S^{32}, {}_{9}F^{19}, {}_{0}n^{1}$.

Nuclear reactions are given as $(i)$ $(n, p) {}_{16}S^{32} \to {}_{15}P^{32}$,$(ii)$ $(p, \alpha) {}_{8}O^{16} \to {}_{9}F^{19}$,$(iii)$ ${}_{7}N^{14} + ? \to {}_{6}C^{14} + {}_{1}H^{1}$. The missing particles or nuclides in these reactions are respectively:

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