Atoms & Nuclei

JEE Main level, single correct, medium

182 of 239

Question 182 of 239

Two nuclei of mass number 33 combine with another nucleus of mass number 44 to yield a nucleus of mass number 1010. If the binding energy per nucleon for the mass numbers 33, 44 and 1010 are 5.65.6 MeV, 7.47.4 MeV and 6.16.1 MeV, respectively, then in the process, ΔMc2=\Delta Mc^2 = _____ MeV.

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Answer: (C) 2.22.2

The nuclear reaction is given by:

2X3+Y4→Z102 X^{3} + Y^{4} \rightarrow Z^{10}

The total binding energy of the reactants is the sum of the binding energies of the two nuclei of mass number 33 and the nucleus of mass number 44.

Binding energy of reactants =2×(3×5.6)+(4×7.4)= 2 \times (3 \times 5.6) + (4 \times 7.4)

=33.6+29.6=63.2= 33.6 + 29.6 = 63.2 MeV

The total binding energy of the product nucleus of mass number 1010 is:

Binding energy of product =10×6.1=61.0= 10 \times 6.1 = 61.0 MeV

The change in mass energy ΔMc2\Delta M c^2 is equal to the difference in the total binding energies of the reactants and the product.

ΔMc2=∣63.2−61.0∣=2.2\Delta M c^2 = |63.2 - 61.0| = 2.2 MeV

Answer: 2.22.2

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