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A certain radioactive substance has a half-life of 40 minutes. How much of an original 100 grams substance will remain after 2 hours? solution) m= m0/2t/Tm=100 gms..t=2 hours=2*60=120 min..T=40 minm=100/2120/40m=100/23m=100/8m=12.5 gms.
law of successive diintegration
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half life = 40minutes =2/3 hourst(1/2)=(ln2 /k)then k=(ln2/t(1/2)) = ln2/(2/3) = 3/2*(ln2)origianl substance y0= 1000 gramsy=y0e-kty= 1000*(e-(3/2(ln2))*2 ) = 1000/8 = 125 grams
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