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The binding energies per nucleon for a deuteron and an `alpha-`particle are `x_1` and `x_2` respectively. What will be the energy `Q` released in the following reaction ? `._1H^2 + ._1H^2 rarr ._2He^4 + Q`.A. `(x_(1)+x_(2))`B. `(x_(2)-x_(1))`C. `4(x_(1)-x_(2))`D. `4(x_(2)-x_(1))` |
Answer» Correct Answer - D `(4x1.0089+5x1.009)-9.012=Deltam` binding energy `=Deltamxxc^(2)` Bindng energy /nucleon `=("Binding energy")//9=6.72 MeV` |
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