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JEE MAIN_2026
04-04-26_S2
Question
Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass numbers 3,4 and 10 are 5.6 MeV , 7.4 MeV and 6.1 MeV , respectively, then in the process, $\Delta \mathrm{Mc}^2=$ $\_\_\_\_$ MeV .
Select the correct option:
A
6.9
B
7.9
C
2.2
D
4.3
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \Delta \mathrm{mc}^2=[2 \times 3 \times 5.6+4 \times(7.4)-6.1 \times 10] \\ & =2.2 \mathrm{MeV}\end{aligned}$
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