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Question

A nuclear fission is given below:
A240B100+C140+Q(energy)
Let binding energy per nucleon of nucleus A,B and C is 7.6MeV, 8.1MeV and 8.1MeV respectively. Value of Q is : (Approximately)

A
120MeV
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B
220MeV
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C
20MeV
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D
240MeV
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Solution

The correct option is A 120MeV
Q = Related every = nB¯EB+nC¯ECnA¯EA
Q=1ω×8.1MeV+140×8.1MeV
240×7.6MeV
Q=1944MeV1824MeV=120MeV

1188817_1105410_ans_08e40f8f47774ab78d87edf5b67e3a4d.jpg

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