(A) The radioactive decay follows the law $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$, where $T_{1/2}$ is the half-life.Given $N(0) = 1600$ an

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(A) The radioactive decay follows the law $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$, where $T_{1/2}$ is the half-life.Given $N(0) = 1600$ and $N(8) = 100$.$100 = 1600 \left( \frac{1}{2} \right)^{8/T_{1/2}}$$\frac{1}{16} = \left( \frac{1}{2} \right)^{8/T_{1/2}}$$\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{8/T_{1/2}}$$4 = \frac{8}{T_{1/2}} \implies T_{1/2} = 2 \, s$.Now, at $t = 6 \, s$, the number of half-lives elapsed is $n = \frac{6}{2} = 3$.The count rate is $N(6) = 1600 \times \left( \frac{1}{2} \right)^3 = \frac{1600}{8} = 200 \, \text{counts per second}$.

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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Using a nuclear counter, the count rate of emitted particles from a radioactive source is measured. At $t = 0$, it was $1600$ counts per second, and at $t = 8 \, s$, it was $100$ counts per second. The count rate observed, in counts per second, at $t = 6 \, s$ is close to:

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Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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Using a nuclear counter, the count rate of emitted particles from a radioactive source is measured. At $t = 0$, it was $1600$ counts per second, and at $t = 8 \, s$, it was $100$ counts per second. The count rate observed, in counts per second, at $t = 6 \, s$ is close to:

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$A$ radioactive sample decays by $\beta$-emission. In the first $2 \ s$,$n$ $\beta$-particles are emitted,and in the next $2 \ s$,$0.25n$ $\beta$-particles are emitted. The half-life of the radioactive nuclei is ...... $s$.

The rate of disintegration of a fixed quantity of a radioactive element can be increased by

$A$ radioactive material whose half-life period is $2$ years weighs $1 \,g$ and is stored in the laboratory for $4$ years. Then the amount of remaining radioactive material is (in $\,g$)

Two radioactive materials $X_1$ and $X_2$ have decay constants $10 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $1 / e$ after a time:

The half-life of $Au^{198}$ is $2.7 \, days$. The activity of $1.50 \, mg$ of $Au^{198}$ if its atomic weight is $198 \, g \, mol^{-1}$ is ....... $Ci$ $(N_A = 6 \times 10^{23} \, mol^{-1})$.

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