In the following nuclear reaction,the other p | vedclass

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(C) To find the missing product,we balance the mass numbers and atomic numbers on both sides of the reaction.Mass number balance: $130 + 2 = 131 + A \

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One neutron

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(C) To find the missing product,we balance the mass numbers and atomic numbers on both sides of the reaction.Mass number balance: $130 + 2 = 131 + A \implies 132 = 131 + A \implies A = 1$.Atomic number balance: $52 + 1 = 53 + Z \implies 53 = 53 + Z \implies Z = 0$.The particle with mass number $1$ and atomic number $0$ is a neutron,represented as $_{0}n^{1}$.Thus,the reaction is: $_{52}Te^{130} + _{1}H^{2} 	o _{53}I^{131} + _{0}n^{1}$.

In the following nuclear reaction,the other product is
$_{52}Te^{130} + _{1}H^{2} \xrightarrow{\quad} _{53}I^{131} + ?$

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In the following nuclear reaction,the other product is$_{52}Te^{130} + _{1}H^{2} \xrightarrow{\quad} _{53}I^{131} + ?$