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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    As a result of radioactive decay \[_{92}{{U}^{238}}\] is converted into \[_{91}P{{a}^{234}}\]. The particles emitted during this decay are

    A)  a proton and a neutron

    B)  a proton and two \[\alpha \]-particles

    C)  an\[\alpha -\]particle and a \[\beta \]-particle

    D)  two p-particles and a proton

    Correct Answer: C

    Solution :

    Decrease in mass number\[=238-234=4\] No. of\[\alpha -\]particles emitted\[=\frac{4}{4}=1\] Decrease in atomic number due to emission of 1\[\alpha -\]particle is 2, but now atomic number is 91, hence number of\[\beta -\]particles\[~=91+2-92=1\].


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