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

Option: 1

{ }_{84} \mathrm{X}^{228} \quad


Option: 2

\quad{ }_{86} \mathrm{X}^{222} \quad


Option: 3

\quad{ }_{83} \mathrm{X}^{216} \quad


Option: 4

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


Answers (1)

best_answer

Decrease in mass number due to emission of 3 \alpha -particles and a \beta -particle =3 \times 4=12

Decrease in charge number in the process =3 \times 2-1=5

\therefore \text { For the resulting element, }

\begin{aligned} & A=228-12=216 \\ & Z=88-5=83 \end{aligned}

Thus, final isotope is { }_{83} \mathrm{X}^{216} . 

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Sayak

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