Activities of three radioactive substances A,B and C are represented by the curves A,B and C in the figure . Then their half lives T12A:T12B:T12C: are in the ratio :

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2 : 1 : 3

4 : 3 : 1

3 : 2 : 1

2 : 1 : 1

answer is A.

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Detailed Solution

Activity, R=R02-tth ⇒lnR=lnR0 -ln2thtComparing with straight line, y=c+mx , slope=m=-ln2thmA=6/10 from slopemB=6/5mC=2/5Since, slope ∝1half life⇒TA:TB:TC=106:56:52 = 20:10:30=2:1:3

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