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Nuclei

Question
CBSEENPH12040149

When the radioactive isotope 22688 decays in a series by the emission of three alpha (α) and a beta (β) particle, the isotope X which remains undecay is

  • X21483

  • X21883

  • X22083

  • X22383

Solution

A.

X21483

Due to the emission of 3α particles, the mass number is reduced by 12 while the atomic number is decreased by 6. Due to the emission of β particle, the atomic number is increased by one while there is no change in mass number. Now, resulting mass will be decreased by 12 and atomic number is decreased by 5. So, isotopes X is 83X214