After the emission of one $\alpha$-particle followed by one $\beta$-particle from the atom of $_{92}X^{238}$,the number of neutrons in the atom will be

  • A
    $142$
  • B
    $146$
  • C
    $144$
  • D
    $143$

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After the emission of an $\alpha$-particle from the atom $_{92}X^{238}$,the number of neutrons in the resulting atom will be:

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