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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
    Consider the following nuclear reactions, \[_{92}^{238}M\to \,_{y}^{x}N+2\,_{2}^{4}He\]                 \[_{y}^{x}N\to \,_{B}^{A}L+2{{\beta }^{+}}\]                 The number of neutrons in the element L is        [AIEEE 2004]

    A)                 140        

    B)                 144

    C)                 142        

    D)                 146

    Correct Answer: B

    Solution :

               \[_{92}{{M}^{238}}\xrightarrow{{}}{{\,}_{y}}{{N}^{x}}+2{{\,}_{2}}H{{e}^{4}}\]                    \[_{y}{{N}^{x}}\xrightarrow{{}}{{\,}_{B}}{{L}^{A}}+2{{\beta }^{+}}\]                    \[_{y}{{N}^{x}}{{=}_{(92-2\times 2)}}{{N}^{(238-4\times 2)}}={{\,}_{88}}{{N}^{230}}\]                    \[_{88}{{N}^{230}}\xrightarrow{2{{\beta }^{+}}}{{\,}_{(88-2)}}{{L}^{(230)}}={{\,}_{86}}{{L}^{230}}\]                    Total no of neutrons in \[_{90}{{L}^{330}}\]                                 \[230-86=144\]


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