Question

${}_7^{14}N{+}_0^{1}n{\to }_6^{14}C{+}_1^{1}H$ is written as

• A. ${}_7^{14}N(n,e{)}_1^{1}H$
• B. ${}_7^{14}N(p,n{)}_6^{14}C$
• C. ${}_7^{14}N(n,p{)}_6^{14}C$
• D. ${}_6^{14}C(p,n{)}_7^{14}N$

Answer 1

The correct option is C ${}_7^{14}N(n,p{)}_6^{14}C$

${}_7^{14}N{+}_0^{1}n{\to }_6^{14}C{+}_1^{1}H$

This reaction can be represented as :

${}_7^{14}N(n,p{)}_6^{14}C$ since one neutron and ${}_7^{14}N$ are in the reactant side and one proton (hydrogen) and ${}_6^{12}C$ are form in the product.