When a radioactive isotope${\mathrm{ }}_{88}{\mathrm{R}}^{228}$decay in series by the emission of $3\alpha$-particles and $\mathrm{\beta }$-particle, the isotope finally formed is

• A

  ${\mathrm{ }}_{84}{\mathrm{X}}^{228}$
• B

  ${\mathrm{ }}_{86}{\mathrm{X}}^{222}$
• Correct answer
  C

  ${\mathrm{ }}_{83}{\mathrm{X}}^{216}$
• D

  ${\mathrm{ }}_{83}{\mathrm{X}}^{215}$

Hence, the correct answer is option: 'C'.