Q. Calculate the energy released when three $\alpha - particles\ $combined to from a $\ ^{12}C$ nucleus , the mass defect is (atomic mass of $\ _{2}He^{4}$ is 4.002603 u)
A) 0.007809 u
B) 0.002603 u
C) 4.002603 u
D) 0.5 u
Explanation
(a) Mass defect
$\mathrm{\Delta}m = Total\ mass\ of\ \alpha - particles - mass\ of\ \ ^{12}C\ nucleus$
$= 3 \times \ 4.002603 - 12$
$= 12.007809 - 12$
$= 0.007809\ unit$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.