Respuesta :
Answer:
The nuclear equation is give as [tex]\ce{^{235}_{92}U} + \ce{^{1}_{0}n} \rightarrow \ce{^{134}_{54}Xe} + \ce{^{100}_{38}Sr} + 2 (\ce{^{1}_{0}n})[/tex]
The number of neutrons produced = 2
Explanation:
Since the reaction is a neutron induced fission of Uranium, a certain amount of neutron will be emitted with Xe - 34 and Sr - 100
Therefore the nuclear equation is given as:
[tex]\ce{^{235}_{92}U} + \ce{^{1}_{0}n} \rightarrow \ce{^{134}_{54}Xe} + \ce{^{100}_{38}Sr} + y (\ce{^{1}_{0}n})[/tex]
To get the value of y:
Sum of mass numbers at the reactant = Sum of mass numbers at the product
235 + 1 = 134 + 100 + y
y = 236 - 234
y = 2
Therefore, the nuclear equation above becomes:
[tex]\ce{^{235}_{92}U} + \ce{^{1}_{0}n} \rightarrow \ce{^{134}_{54}Xe} + \ce{^{100}_{38}Sr} + 2 (\ce{^{1}_{0}n})[/tex]
As seen from the equation above, the number of neutrons produced in the reaction = 2
