An element $_{96}X^{227}$ emits $4\alpha$ and $5\beta$ particles to form a new element $Y$. The atomic number and mass number of $Y$ are:

  • A
    $93; 211$
  • B
    $211; 93$
  • C
    $212; 88$
  • D
    $88; 211$

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In the sequence of the following nuclear reactions $_{92}X^{238}$ $\xrightarrow{-\alpha} Y$ $\xrightarrow{-\beta} Z$ $\xrightarrow{-\beta} L$ $\xrightarrow{-n\alpha} _{84}M^{218}$. The value of $n$ will be:

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