A nuclear fission is given below A240→B100+C140+Q(energy).
Let binding energy per nucleon of nuclei A,B and C be 7.6 MeV,8.1 MeV and 8.1 MeV respectively. Value of Q is : (Approximately)
A
20MeV
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B
220MeV
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C
120MeV
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D
240MeV
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Solution
The correct option is C120MeV Binding Energy of A=240×7.6=1824 MeV
Binding Energy of B=100×8.1=810 MeV
Binding Energy of C=140×8.1=1134 MeV
So Q=(810+1134)MeV−1824 MeV=120 MeV