Kinetic energy of the emitted $\alpha-$ particle in the $\alpha-$ decay of ${}_{88}^{226}Ra$ will be,.......... $MeV$ (where $m_{\alpha} = 4.00260 \, u$,$m({}_{88}^{226}Ra) = 226.02540 \, u$ and $m({}_{86}^{222}Rn) = 222.01750 \, u$).

  • A
    $8.93$
  • B
    $4.93$
  • C
    $8.77$
  • D
    $4.85$

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