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Two radioactive elements R and S disintegrate as `RtoP+alpha,lamda_(R)=4.5xx10^(-3)"years"^(-1)` `StoQ+betal,lamda_(S)=3xx10^(-3)"years"^(-1)` Starting with number of atoms of R and S in the ratio of 2:1 this ratio afte4r the lapse of three half lives of R will beA. `3:2`B. `1:3`C. `1:1`D. `2:1` |
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Answer» Correct Answer - C `(lamda_(R))/(lamda_(S))=1.5` So, the rate of disintegration of R will be 1.5 times that of S. Thus, the half life of S will be 1.5 times that of R. So, two half lives of S will be equal to the three half -lives of R `(N_(R))/(N_(S))=(0.25)/(0.25)=1` |
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