After 3 hours,only 0.25 ,mg of a pure radioac | vedclass

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(B) Given:Initial mass $N_0 = 2 \,mg$Final mass $N = 0.25 \,mg$Total time $t = 3 \,hr$The radioactive decay law is given by $N = N_0 \left(\frac{1}{2}

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(B) Given:Initial mass $N_0 = 2 \,mg$Final mass $N = 0.25 \,mg$Total time $t = 3 \,hr$The radioactive decay law is given by $N = N_0 \left(\frac{1}{2}ight)^n$,where $n$ is the number of half-lives.Substituting the values:$0.25 = 2 	imes \left(\frac{1}{2}ight)^n$$\frac{0.25}{2} = \left(\frac{1}{2}ight)^n$$\frac{1}{8} = \left(\frac{1}{2}ight)^n$$\left(\frac{1}{2}ight)^3 = \left(\frac{1}{2}ight)^n$Therefore,$n = 3$.Since $n = \frac{t}{T_{1/2}}$,where $T_{1/2}$ is the half-life:$3 = \frac{3 \,hr}{T_{1/2}}$$T_{1/2} = 1 \,hr$.

After $3$ hours,only $0.25 \,mg$ of a pure radioactive material is left. If the initial mass was $2 \,mg$,then the half-life of the substance is ...... $hr$.

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