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BHU PMT BHU PMT (Screening) Solved Paper-2011

  • question_answer
    \[_{92}{{U}^{238}}\]on absorbing a neutron goes over to\[_{92}{{U}^{239}}\]. This nucleus emits an electron to go over to neptunium which on further emitting an electron goes over to plutonium. The resulting plutonium can be expressed as

    A)  \[_{94}P{{u}^{239}}\]                   

    B)  \[_{90}P{{u}^{219}}\]

    C)  \[_{93}P{{U}^{240}}\]                  

    D) \[_{92}P{{U}^{240}}\]

    Correct Answer: A

    Solution :

                     The reaction can be shown as \[_{92}{{U}^{238}}{{+}_{0}}{{n}^{1}}{{\xrightarrow{{}}}_{92}}{{U}^{239}}\xrightarrow[{}]{{}}\] \[_{93}N{{e}^{239}}{{+}_{-1}}{{e}^{0}}{{\xrightarrow{{}}}_{-1}}{{e}^{0}}{{+}_{94}}P{{u}^{239}}\]


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