When a radioactive isotope {}_{88}^{228}Ra de | vedclass

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(C) The initial isotope is ${}_{88}^{228}Ra$.An alpha particle emission $({}_{2}^{4}He)$ reduces the mass number $(A)$ by $4$ and the atomic number $(

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${}_{83}^{216}X$

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(C) The initial isotope is ${}_{88}^{228}Ra$.An alpha particle emission $({}_{2}^{4}He)$ reduces the mass number $(A)$ by $4$ and the atomic number $(Z)$ by $2$.$A$ beta particle emission $({}_{-1}^{0}e)$ increases the atomic number $(Z)$ by $1$ while the mass number $(A)$ remains unchanged.After emitting $3$ alpha particles: New $A = 228 - (3 	imes 4) = 228 - 12 = 216$.New $Z = 88 - (3 	imes 2) = 88 - 6 = 82$.After emitting $1$ beta particle:Final $A = 216$ (remains unchanged).Final $Z = 82 + 1 = 83$.Therefore,the final isotope is ${}_{83}^{216}X$.

When a radioactive isotope ${}_{88}^{228}Ra$ decays in series by the emission of $3$ alpha particles and $1$ beta particle,the isotope finally formed is

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