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The binding energy per nucleon of `._(3)^(7) Li` and `._(2)^(4)He` nuclei are `5.60` MeV and `7.06` MeV, respectively. In the nuclear reaction `._(3)^(7)Li+._(1)^(1)H rarr ._(2)^(4)He+._(2)^(4)He+Q`, the value of energy `Q` released isA. `19.6` Me VB. `-2.4` Me VC. `8.4` Me VD. `17.3` Me V |
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Answer» Correct Answer - D The binding energy for `""_(1)H^(1)` is around zero and also not given in the question so we can igonor it `Q=2(4xx7.06)-7xx(5.60)=(8xx7.06)-(7xx5.60)` `=56.48-39.2=17.28MeV~=17.3 MeV` |
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