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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    An isotope \[_{Y}^{X}A\] undergoes a series of m alpha and n beta disintegration to form a stable isotope\[_{Y-10}^{X-32}B\]. The values of m and n are respectively

    A)  \[6\] and \[8\]                 

    B)  \[8\] and \[10\]

    C)  \[5\] and \[8\]                 

    D)  \[8\] and \[6\]

    Correct Answer: D

    Solution :

    \[_{Y}{{A}^{X}}\xrightarrow[\beta =n]{\alpha =m}\,_{Y-10}^{X-32}B\] given, \[\alpha -particle=m,\]\[\beta -particle=n\]          difference between                 \[m(\alpha -particle)=\frac{at.\,\,wt.\,\,of\,\,A\,and\,B}{4}\]                 \[=\frac{32}{4}=8\alpha particles\] \[n(\beta -particle)=2\times \alpha -particle-difference\] between at. no. of A and B                 \[=2\times 8-10\]                 \[=16-10\]                 \[=6\beta -particles\] So, the value of \[m=8,\] the value of \[n=6.\].


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