Q. If the binding energy of the deutrium is $2.23\ MeV$. The mass defect given in $a.m.u.$ is
A) $- 0.0024$
B) $- 0.0012$
C) 0.0012
D) 0.0024
Explanation
(d) Mass defect $\mathrm{\Delta}m = \frac{2.23}{931} = 0.0024\ amu$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.