Atoms and Nuclei: JEE Main Physics Question with Solution
A nucleus with mass number 242 and binding energy per nucleon as 7.6 MeV breaks into fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV, the total gain in binding energy is MeV
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Hint 1 of 4
How is the total binding energy of a nucleus related to its mass number A and binding energy per nucleon (BE/A)?
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Correct answer
The total gain in binding energy is 121 MeV.
Option analysis
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Reviewed route
Solution
StepWorking
01given
Parent nucleus mass number A=242 with BE/nucleon=7.6 MeV. It breaks into two fragments each of mass number A1=A2=121 with BE/nucleon=8.1 MeV.
02find
Total gain in binding energy ΔBE in MeV.
03visualise
A heavy parent nucleus splits symmetrically into two identical daughter fragments: X(242)→Y(121)+Y(121). Total nucleons remain conserved at 242.
04strategise
Gain in BE=BEfinal−BEinitial=A×(BE/nucleon)final−A×(BE/nucleon)initial=A×Δ(BE/nucleon).
05execute
ΔBE=242×(8.1−7.6)=242×0.5=121 MeV.
✓verify
Units are in MeV. A gain of 0.5 MeV/nucleon across 242 nucleons gives 121 MeV, which is positive and physically typical for a fission reaction.
✓ Source and academic review↓
Question type
Numerical
Exam relevance
JEE Main · Physics
Concepts assessed
Physics
Academic status
Reviewed by official_key
Source
pyq
Editorial review
7 September 2026
Quick checks
Students also ask
Why do we multiply by 242 instead of summing 121 for just one fragment?
The question asks for the total gain across the entire reaction. The reaction produces two fragments of mass number 121. Thus, the total number of nucleons is 121+121=242.