The nuclear reaction _{29}^{63}Cu + _{2}^{4}H | vedclass

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Question

(A) The given reaction is $_{29}^{63}Cu + _{2}^{4}He 	o _{17}^{37}Cl + 14_{1}^{1}H + 16_{0}^{1}n$. In this reaction,a high-energy alpha particle $(_{

Vedclass · Accepted Answer

Spallation reaction

Vedclass · Answer

(A) The given reaction is $_{29}^{63}Cu + _{2}^{4}He 	o _{17}^{37}Cl + 14_{1}^{1}H + 16_{0}^{1}n$. In this reaction,a high-energy alpha particle $(_{2}^{4}He)$ strikes a copper nucleus,causing it to break into several smaller fragments including a chlorine nucleus,protons,and neutrons. This process,where a heavy nucleus is broken into many smaller fragments by high-energy bombardment,is known as a spallation reaction. Therefore,the correct option is $A$.

The nuclear reaction $_{29}^{63}Cu + _{2}^{4}He \to _{17}^{37}Cl + 14_{1}^{1}H + 16_{0}^{1}n$ is referred to as

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