The half-life of a radioactive nucleus is 50 | vedclass

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(B) According to the radioactive decay law,$N = N_0 e^{-\lambda t}$,where $N_0$ is the initial number of nuclei and $N$ is the number of undecayed nuc

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$50$

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(B) According to the radioactive decay law,$N = N_0 e^{-\lambda t}$,where $N_0$ is the initial number of nuclei and $N$ is the number of undecayed nuclei at time $t$.At time $t_2$,$2/3$ of the sample has decayed,so the remaining amount is $N = N_0 - (2/3)N_0 = (1/3)N_0$.Thus,$(1/3)N_0 = N_0 e^{-\lambda t_2} \Rightarrow e^{-\lambda t_2} = 1/3$ ...... $(i)$At time $t_1$,$1/3$ of the sample has decayed,so the remaining amount is $N = N_0 - (1/3)N_0 = (2/3)N_0$.Thus,$(2/3)N_0 = N_0 e^{-\lambda t_1} \Rightarrow e^{-\lambda t_1} = 2/3$ ...... $(ii)$Dividing $(i)$ by $(ii)$,we get: $\frac{e^{-\lambda t_2}}{e^{-\lambda t_1}} = \frac{1/3}{2/3} = 1/2$.This simplifies to $e^{-\lambda(t_2 - t_1)} = 1/2$,or $e^{\lambda(t_2 - t_1)} = 2$.Taking the natural logarithm on both sides: $\lambda(t_2 - t_1) = \ln 2$.Since $\lambda = \frac{\ln 2}{T_{1/2}}$,we have $(t_2 - t_1) = \frac{\ln 2}{\lambda} = T_{1/2}$.Given $T_{1/2} = 50$ days,the time interval $(t_2 - t_1) = 50$ days.

The half-life of a radioactive nucleus is $50$ days. The time interval $(t_2 - t_1)$ between the time $t_2$ when $2/3$ of it has decayed and the time $t_1$ when $1/3$ of it has decayed is ...... days.

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