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Question

A nuclear fission is given below
A240B100+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
20 MeV
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B
220 MeV
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C
120 MeV
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D
240MeV
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Solution

The correct option is C 120 MeV
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)MeV1824 MeV=120 MeV

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