AIIMS2018Physics-Nuclear Physics

AIIMS 2018 Physics Energy Calculations MCQ Question

Type: MCQ-numerical-Medium-Class 12

If the binding energy per nucleon in 37Li_3^7\text{Li} and 24He_2^4\text{He} nuclei are 5.60 MeV5.60\text{ MeV} and 7.06 MeV7.06\text{ MeV} respectively, then in the reaction:

p+37Li2 24He\text{p} + {}_3^7\text{Li} \rightarrow 2\ {}_2^4\text{He}

The energy released (or kinetic energy gain) must be:

A

28.24 MeV

B

17.28 MeV

C

1.46 MeV

D

39.2 MeV

Correct Answer

Option B

Detailed Explanation

To solve for the energy of the proton EE, we rearrange the equation E+7×5.6=2×[4×7.06]E + 7 \times 5.6 = 2 \times [4 \times 7.06]. Calculating the right side, we find 2×[4×7.06]=56.482 \times [4 \times 7.06] = 56.48, and substituting 7×5.6=39.27 \times 5.6 = 39.2 gives E+39.2=56.48E + 39.2 = 56.48. Solving for EE results in E=56.4839.2=17.28 MeVE = 56.48 - 39.2 = 17.28 \text{ MeV}, confirming that option B is correct. Other options are not applicable as they do not provide relevant values or solutions to the problem.

Found an issue with this question?