A parent nucleus X undergoes alpha-undergoes alpha-decay with a half-life of 75000 years. The daughter nucleus Y undergoes beta-decay with a half-life of 9 months. In a particular sample, it is found that the rate of emission of beta-particles is nearly constant (over several months) at 10^(7)//"hour". What will be the number of alpha-particles emitted in an hour ?

A parent nucleus X undergoes alpha-undergoes alpha-decay with a half-l

1.	A parent nucleus X undergoes alpha-undergoes alpha-decay with a half-life of 75000 years. The daughter nucleus Y undergoes beta-decay with a half-life of 9 months. In a particular sample, it is found that the rate of emission of beta-particles is nearly constant (over several months) at 10^(7)//"hour". What will be the number of alpha-particles emitted in an hour ?
Answer» `10^(2)` `10^(7)` `10^(12)` `10^(14)` Solution :As we know that RADIOACTIVE decay is a spontaneous process. So, emission of alpha and beta particles can occur simultaneously. `therefore` From the law of radioactive decay, `(-(dN)/(dt))_(alpha-"particle") = lambda_(alpha) N_(alpha) and (-(dN)/(dt))_(beta-"particle") = lambda_(beta) N_(beta)` Now, `lambda_(alpha)N_(alpha) = lambda_(beta) N_(beta)` (At steady STATE) `rArr""(-(dN)/(dt))_(alpha-"particle") = (-(dN)/(dt))_(beta-"particle")` `rArr""(-(dN)/(dt))_(alpha-"particle") = 10^(7)//"hour"`

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