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(C) Let $N_0$ be the initial number of radioactive nuclei and $T$ be the half-life.The number of nuclei decayed in time $t$ is given by $\Delta N = N_

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(C) Let $N_0$ be the initial number of radioactive nuclei and $T$ be the half-life.The number of nuclei decayed in time $t$ is given by $\Delta N = N_0(1 - 2^{-t/T})$.In the first $2 \ s$,the number of decayed nuclei is $n = N_0(1 - 2^{-2/T})$.In the next $2 \ s$ (from $t=2$ to $t=4$),the number of decayed nuclei is $0.25n = N_0(1 - 2^{-4/T}) - N_0(1 - 2^{-2/T}) = N_0(2^{-2/T} - 2^{-4/T})$.Dividing the two equations: $\frac{n}{0.25n} = \frac{1 - 2^{-2/T}}{2^{-2/T}(1 - 2^{-2/T})}$.$4 = \frac{1}{2^{-2/T}} = 2^{2/T}$.Since $4 = 2^2$,we have $2 = 2/T$,which gives $T = 1 \ s$.

$A$ 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 ...... $s$.

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