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The count rate of a radioactive sample falls from 4.0xx10^6 s^(-1) to 1.0xx10^6 s^(-1) in 20 hours. What will be the count rate after 100 hours from beginning ? |
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Answer» `3.91xx10^3 s^(-1)` t=20 hours, t'=100 hours As,`R/R_0(1/2)^n` or `(1.0xx10^6)/(4.0xx10^6)=(1/2)^n` or `(1/2)^n = (1/2)^2 THEREFORE n=2` `T_(1//2)=t/n=20/2`=10 hours Now, `n'="t'"/T_(1//2)=100/10`=10 `therefore R'=R_0(1/2)^(n') =4.0xx10^6 (1/2)^10 = 3.91xx10^3 s^(-1)` |
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