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J & K CET Engineering J and K - CET Engineering Solved Paper-2012

  • question_answer
    In a nuclear fusion reaction, two nuclei, A and B, fuse to produce a nucleus C, releasing an amount of energy \[\Delta E\] in the process. If the mass defects of the three nuclei are \[\Delta {{M}_{A}},\] \[\Delta {{M}_{B}}\] and am(; respectively, then which of the following relations holds?  Here is the speed of light

    A)  \[\Delta {{M}_{A}}+\Delta {{M}_{B}}=\Delta {{M}_{C}}-\Delta E/{{c}^{2}}\]

    B)  \[\Delta {{M}_{A}}+\Delta {{M}_{B}}=\Delta {{M}_{C}}+\Delta E/{{c}^{2}}\]

    C)  \[\Delta {{M}_{A}}-\Delta {{M}_{B}}=\Delta {{M}_{C}}-\Delta E/{{c}^{2}}\]

    D)  \[\Delta {{M}_{A}}-\Delta {{M}_{B}}=\Delta {{M}_{C}}+\Delta E/{{c}^{2}}\]

    Correct Answer: A

    Solution :

    Binding energy of nuclei \[A=\Delta {{M}_{A}}{{C}^{2}}\] Binding energy of nuclei \[B=\Delta {{M}_{B}}{{C}^{2}}\] Binding energy of nuclei  \[C=\Delta {{M}_{C}}{{C}^{2}}\] According to question \[A+B\xrightarrow{{}}C\] So, released energy \[\Delta E=BE\] of C (\[BE\]of A and B) \[=\Delta {{M}_{C}}{{C}^{2}}-(\Delta {{M}_{A}}{{C}^{2}}+\Delta {{M}_{B}}{{C}^{2}})\] or \[\frac{\Delta E}{{{c}^{2}}}=\Delta {{M}_{C}}-(\Delta {{M}_{A}}+\Delta {{M}_{B}})\] \[\Delta {{M}_{A}}+\Delta {{M}_{B}}=\Delta {{M}_{C}}-\frac{\Delta E}{{{c}^{2}}}\]


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