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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Self Evaluation Test - Nuclei

  • question_answer
    The element curium \[_{96}C{{m}^{248}}\] has a mean life of \[{{10}^{13}}s.\]. Its primary decay modes are spontaneous fission and \[\alpha \]-decay, the former with a probability of \[8%\] and the later with a probability of \[92%\]. Each fission releases \[200\text{ }MeV\] of energy. The masses involved in a decay are as follows: \[_{96}C{{m}^{248}}=248.072220u,\]\[_{94}P{{u}^{244}}=244.064100\,u\] and \[_{2}H{{e}^{4}}=4.002603u\] Calculate the power output from a sample of \[{{10}^{20}}Cm\] atoms. \[(1\,u=931\,\,MeV/{{c}^{2}}).\]

    A)  \[1.6\times {{10}^{-5}}W\]

    B)  \[2.6\times {{10}^{-3}}W\]

    C)  \[3.3\times {{10}^{-5}}W\] 

    D)  \[5.1\times {{10}^{-3}}W\]

    Correct Answer: C

    Solution :

    [c] The total energy released E= Energy released in fission process + energy released in \[\alpha \]- decay process             \[={{N}_{F}}\times 200+{{N}_{\alpha }}\times (0.005517\times 931)\] \[=\left( \frac{8}{100}\times {{10}^{20}} \right)\times 200+\left( \frac{92}{100}\times {{10}^{20}} \right)\] \[\times (0.005517\times 931)\]             \[=20.725\times {{10}^{20}}MeV\] Power output \[P=E/t\] \[=\frac{20.725\times {{10}^{20}}\times 1.6\times {{10}^{-13}}}{{{10}^{13}}}\] \[=3.3\times {{10}^{-5}}W.\]


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