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(B) The radioactive sample emits $64$ times the non-hazardous limit,so the ratio of current activity to the limit is $N/N_0 = 1/64$.Using the radioact

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$12$

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(B) The radioactive sample emits $64$ times the non-hazardous limit,so the ratio of current activity to the limit is $N/N_0 = 1/64$.Using the radioactive decay formula: $N = N_0(1/2)^n$,where $n$ is the number of half-lives.$1/64 = (1/2)^n$ $\Rightarrow (1/2)^6 = (1/2)^n$ $\Rightarrow n = 6$.Given the half-life $t_{1/2} = 2 \ hr$,the total time $T = n 	imes t_{1/2} = 6 	imes 2 = 12 \ hr$.Thus,the sample becomes non-hazardous after $12 \ hr$.

$A$ radioactive sample is emitting $64$ times more radiation than the non-hazardous limit. If its half-life is $2 \ hr$,after what time will it become non-hazardous? $(hr)$

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