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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    A radioactive element \[_{90}{{X}^{238}}\] decays into \[_{83}{{\Upsilon }^{222}}.\] The number of P-particle emitted are:

    A)  1                            

    B)         2                            

    C)  4                            

    D)         6

    Correct Answer: A

    Solution :

    \[\alpha \] -particle are emitted\[=\frac{238-222}{4}=4\] The atomic number is decreased \[90-4\times 2=82\] As atomic number of \[{}_{83}{{Y}^{222}},\] so atomic number is increased by 1, therefore one \[\beta \]-particle is emitted.


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