A radioactive material has a half-life of 8 y | vedclass

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(D) The radioactive decay law is given by the formula: $\frac{A}{A_0} = (\frac{1}{2})^{t/T_{1/2}}$, where $A$ is the final activity, $A_0$ is the init

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$24$

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(D) The radioactive decay law is given by the formula: $\frac{A}{A_0} = (\frac{1}{2})^{t/T_{1/2}}$, where $A$ is the final activity, $A_0$ is the initial activity, $t$ is the time elapsed, and $T_{1/2}$ is the half-life.Given that the activity decreases to $1/8$ of its original value, we have $\frac{A}{A_0} = 1/8$.The half-life $T_{1/2} = 8$ years.Substituting these values into the formula: $1/8 = (\frac{1}{2})^{t/8}$.Since $1/8 = (\frac{1}{2})^3$, we can write: $(\frac{1}{2})^3 = (\frac{1}{2})^{t/8}$.Comparing the exponents, we get $3 = t/8$.Therefore, $t = 3 	imes 8 = 24$ years.

$A$ radioactive material has a half-life of $8$ years. The activity of the material will decrease to about $1/8$ of its original value in .......... $years$.

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