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STD 12 - 13. Nuclei — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 13. Nuclei4 Marks
MCQ
Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below:

${ }_{1}^{2} X +{ }_{1}^{2} X ={ }_{2}^{4} Y$

The binding energies per nucleon ${ }_{1}^{2} X$ and ${ }_{2}^{4} Y$ are $1.1\,MeV$ and $7.6\,MeV$ respectively. The energy released in this process is $MeV$.

  • A
    $25$
  • ✓
    $26$
  • C
    $23$
  • D
    $22$
✓

Answer

Correct option: B.
$26$
b
Energy released in the given process $=$ Binding energy of product $-$ Binding energy of reactants

$=7.6 \times 4-(1.1 \times 2) \times 2$

$=30.4-4.4$

$=26\,MeV$

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