In a radioactive substance at t = 0,the numbe | vedclass

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(A) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $N_0$ is the initial number of atoms,$N$ is the remaining n

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$9$

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(A) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} ight)^{t/T}$,where $N_0$ is the initial number of atoms,$N$ is the remaining number of atoms,$t$ is the time elapsed,and $T$ is the half-life period.Given: $N_0 = 8 	imes 10^4$,$N = 1 	imes 10^4$,and $T = 3 \ years$.Substituting the values into the formula:$1 	imes 10^4 = 8 	imes 10^4 \left( \frac{1}{2} ight)^{t/3}$$\frac{1}{8} = \left( \frac{1}{2} ight)^{t/3}$Since $\frac{1}{8} = \left( \frac{1}{2} ight)^3$,we have:$\left( \frac{1}{2} ight)^3 = \left( \frac{1}{2} ight)^{t/3}$Comparing the exponents:$3 = \frac{t}{3}$$t = 9 \ years$.

In a radioactive substance at $t = 0$,the number of atoms is $8 \times 10^4$. Its half-life period is $3 \ years$. The number of atoms $1 \times 10^4$ will remain after an interval of ........... $years$.

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