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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2008

  • question_answer
    The binding energy of deuteron\[_{1}^{2}H\]is 1.112 MeV per nucleon and an \[\alpha \]-particle\[_{2}^{4}He\]has a binding energy of 7.047 MeV per nucleon. Then is the fusion reaction\[_{1}^{2}H+_{1}^{2}H\xrightarrow[{}]{{}}_{2}^{4}H+Q,\]the energy Q released is

    A)  1 MeV          

    B)  11.9 MeV

    C)  23.8 MeV       

    D)  931 MeV

    Correct Answer: C

    Solution :

     Mass of\[_{1}{{H}^{2}}=\]2.01478 amu Mass of\[_{2}H{{e}^{4}}=\]400388 amu Mass of two deuterium \[=2\times 2.01478=4.02956\] Energy equivalent to 2,\[{{H}^{2}}\] \[=4.02956\times 1.112\text{ }MeV=4.48\text{ }MeV\] Energy equivalent to\[_{2}H{{e}^{4}}\] \[=4,00388\times 7.047MeV=28.21\text{ }MeV\] Energy released \[=28.21-4.48=23.73\text{ }MeV\] \[=24\text{ }MeV\]


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