The binding energies of the nuclei A and B are Ea and Eb respectively. Three nuclei of the element B fuse to give one nucleus of element A and an energy Q is released. Then, Ea,Eb,Q are related as :
A
Ea−3Eb,=Q
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B
3Eb−Ea,=Q
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C
Ea+3Eb,=Q
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D
Eb+3Ea,=Q
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Solution
The correct option is AEa−3Eb,=Q BE)A=Ea,BE)B=Eb 3B→A+Q Ea−3Eb=Q