BVP Medical BVP Medical Solved Paper-2015

  • question_answer
    Consider the reaction \[{{v}_{P}}={{v}_{Q}}=0\]. If  \[{{v}_{P}}=0,{{v}_{Q}}=2v\]and \[\eta \]. The energy Q released (in MeV) in this fusion reaction is

    A)  \[\lambda \]

    B)  \[\frac{2\eta }{\lambda }[1-{{e}^{-\lambda t}}]\]

    C)  \[\frac{\eta }{2\lambda }[1-{{e}^{-\lambda t}}]\]

    D)  \[\frac{\eta }{{{\lambda }^{2}}}[1-{{e}^{-{{\lambda }^{2}}t}}]\]

    Correct Answer: B

    Solution :

                    Consider the reaction \[_{1}{{H}^{2}}{{+}_{1}}{{H}^{2}}{{\to }_{2}}H{{e}^{4}}+Q\] \[\Delta m=m{{(}_{2}}H{{e}^{4}})-2m{{(}_{1}}{{H}^{2}})\] \[={{A}_{2}}-2{{A}_{1}}\] Since,    \[Q=\Delta m{{c}^{2}}=({{A}_{2}}-2{{A}_{1}}){{C}^{2}}\]


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