In the following nuclear reaction $_{12}Mg^{24} + _{2}He^{4} \to _{14}Si^{X} + _{0}n^{1}$,the value of $X$ is:

  • A
    $28$
  • B
    $27$
  • C
    $26$
  • D
    $22$

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Reason $(R) :$ The nucleus of mass number $A$ has a radius proportional to $A^{1/3}$.
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