In a radioactive sample, $$1940K$$ nuclei either decay into stable $$2040Ca$$ nuclei with decay constant $$4.5$$×$$1010$$ per year or into stable $$1840Ar$$ nuclei with decay constant $$0.5$$×$$1010$$ per year. Given that in this sample all the stable $$2040Ca$$ and $$1840Ar$$ nuclei are produced by the $$1940K$$ nuclei only. In time t×$$109$$ years, if the ratio of the sum of stable $$2040Ca$$ and $$1840Ar$$ nuclei to the radioactive $$1940K$$ nuclei is $$99$$, the value of t will be [Given: $$In10$$ $$=$$ $$2.3$$ ]

Question

In a radioactive sample,  $$1940K$$ nuclei either decay into stable  $$2040Ca$$ nuclei with decay constant $$4.5$$×$$1010$$ per year or into stable  $$1840Ar$$ nuclei with decay constant $$0.5$$×$$1010$$ per year. Given that in this sample all the stable   $$2040Ca$$ and  $$1840Ar$$ nuclei are produced by the  $$1940K$$  nuclei only. In time t×$$109$$ years, if the ratio of the sum of stable  $$2040Ca$$   and  $$1840Ar$$  nuclei to the radioactive  $$1940K$$  nuclei is $$99$$, the value of t will be [Given: $$In10$$ $$=$$ $$2.3$$ ]

Infinity Learn · Accepted Answer

λ$$eff=$$λ$$1+$$λ$$2=4.5+0.510$$−$$10=5$$×$$10$$−$$10$$  per $$yearN=N0e$$−λt  and let say Nstable $$=N0$$−NNradioactive $$=N$$ Given :Nstable Nradioactive $$=N0$$−$$NN=99$$  ⇒$$N=N0100$$    ⇒$$N0100=N0e$$−λ$$t0$$ ⇒$$102=e$$λ$$t0$$⇒$$2ln10$$λ$$=t0$$∴$$t0=2$$×$$2.35$$×$$10$$−$$10=9.2$$×$$109$$   Years  $$=t$$×$$109$$ YearsSo, t $$=$$ $$9.2$$

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material