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NEET PHYSICSNUCLEIEasy

Question

A nucleus with mass number 240 breaks into fragments each of mass number 120. The binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the binding energy in the process is:

A

804 MeV

B

216 MeV

C

0.9 MeV

D

9.4 MeV

Step-by-Step Solution

  1. Concept: The total binding energy (BEBE) of a nucleus is the product of its mass number (AA) and its binding energy per nucleon (BE/ABE/A). Energy is released (gain in binding energy) when a heavy nucleus splits into lighter, more stable nuclei (fission) with higher binding energy per nucleon .
  2. Calculate Initial Binding Energy (BEiBE_i):
  • Mass number (AA) = 240
  • BE/ABE/A = 7.6 MeV
  • BEi=240×7.6 MeV=1824 MeVBE_i = 240 \times 7.6 \text{ MeV} = 1824 \text{ MeV}.
  1. Calculate Final Binding Energy (BEfBE_f):
  • The nucleus breaks into fragments of mass number 120. Since 240/120=2240/120 = 2, there are two fragments.
  • Total nucleons remain conserved (240).
  • BE/ABE/A (fragments) = 8.5 MeV
  • BEf=240×8.5 MeV=2040 MeVBE_f = 240 \times 8.5 \text{ MeV} = 2040 \text{ MeV}.
  1. Calculate Total Gain in Binding Energy:
  • Gain = BEfBEiBE_f - BE_i
  • Gain = 2040 MeV1824 MeV=216 MeV2040 \text{ MeV} - 1824 \text{ MeV} = 216 \text{ MeV}.
  • Alternatively: Gain=A×(BE/AfinalBE/Ainitial)=240×(8.57.6)=240×0.9=216 MeVGain = A \times (BE/A_{final} - BE/A_{initial}) = 240 \times (8.5 - 7.6) = 240 \times 0.9 = 216 \text{ MeV}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from NUCLEI. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSNUCLEInucleusnumberbreaksfragmentsnumber

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