InterviewSolution
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InterviewSolution
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A sample of `.^(18)F` is used internally as a medical diagnostic tool to look for the effects of the positron decay `(T_(1//2) =110 min)` . How long does it take for `99%` of the `.^(18)F` to decay? |
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Answer» Correct Answer - `12.2 h` Radioactive deacy equation is `N =N_0 e^(- lambda t)=N_0 e^(-1n(2)t//T)" "( because lambda=(In2)/(T))` After decay of `99%` of the initial sample, only `1%` will be left, i.e., `N//N_(0)=1%` `:. (N)/(N_(0))=(1)/(100)=e^(-1n(2)t//T)` If we take the natural logarithm, we have `-1n 100= -1n 2 xx(t)/(T)` Which on solving for t yields `t=(1 n100)/(1n2)xx T =(log 100)/(log2) xxT` `=(2)/(0.3010) xx 10 =731 min =12.2 h`. |
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