Number of nuclei of radioactive substance are $$900$$ and $$810$$ at times $$t=0$$ and $$t=3$$ $$sec$$. Then number of nuclei at time t $$=$$ $$6$$ $$sec$$ will be

Question

Infinity Learn · Accepted Answer

Decay Law , $$N=Noe-$$λt  Case $$1$$ : $$810=900$$ $$e-$$λ$$3$$ ⇒$$e-$$λ$$3$$ $$=$$ $$810900$$ Case $$2$$ : N $$=$$ $$900$$ $$e-$$λ$$6$$ ⇒N $$=$$ $$900$$ $$($$ $$e-$$λ$$3)2$$  ⇒N $$=$$ $$900$$ $$(810900)2$$  ⇒N $$=$$ $$900$$×$$81100=729$$ In first $$3$$ $$sec$$,  $$10$$ percent of the nuclei are decayed from $$900$$ and  then in another $$3$$ $$sec$$, $$10$$ percent nuclei will be decayed from $$810$$ and  hence after $$6$$ $$sec$$ the number of nuclei present is $$729$$.

Number of nuclei of radioactive substance are 900 and 810 at times t=0 and t=3 sec. Then number of nuclei at time t = 6 sec will be

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