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JEE Main & Advanced Chemistry Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन Question Bank Causes of radioactivity and Group displacement law

  • question_answer
    The number of b-particles emitted in radioactive change \[_{92}{{U}^{238}}{{\to }_{82}}P{{b}^{206}}{{+}_{2}}H{{e}^{4}}\] is                [KCET 2000]

    A)                 2             

    B)                 4

    C)                 6             

    D)                 10

    Correct Answer: C

    Solution :

           Suppose the no. of a-particles emitted \[=x\]and                    the no. of \[\beta \]-particles emitted =y. Then                    \[_{92}{{U}^{238}}{{\to }_{82}}P{{b}^{206}}+x_{+2}^{4}\alpha +y\,\,_{-1}^{0}\beta \]                    equating the mass number on both sides, we get                    238 = 206 + 4x + 0 y or 4x = 32, x = 8                    equating the atomic number on both sides, we get                    92 = 82 + 2x ? y                    92 = 82 + 2× 8 ?y                    y = 6                                 Hence 8\[\alpha \]and \[6\beta \] are emitted.


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