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Question: Three $\alpha -$ particles and one $\beta -$ particle decaying takes place in series from an isotope $_{88}R{a^{238}}$. Finally the isotope obtained will be

1) $_{84}{X^{220}}$

2) $_{86}{X^{222}}$

3) $_{83}{X^{224}}$

4) $_{83}{X^{215}}$

Solution:

Explanation: No Explanation

Radioactivity Modern Physics

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