Question

${}_7^{14}N+{}_0^{1}n⟶{}_6^{14}C{+}_1^{1}H$ 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 ${}_7^{14}N(n,p){}_6^{14}C$

Here, by neutron bombardment, proton produces along with daughter element so it is represented as ${}_7^{14}N(n,p){}_6^{14}C$