A radioactive nucleus A with a half-life T de | vedclass

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(C) Let $N_0$ be the initial number of nuclei of $A$ at $t = 0$.At time $t$,the number of nuclei of $A$ remaining is $N_A = N_0 e^{-\lambda t}$.The nu

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$t = T \frac{\log 1.3}{\log 2}$

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(C) Let $N_0$ be the initial number of nuclei of $A$ at $t = 0$.At time $t$,the number of nuclei of $A$ remaining is $N_A = N_0 e^{-\lambda t}$.The number of nuclei of $B$ formed is $N_B = N_0 - N_A = N_0(1 - e^{-\lambda t})$.The ratio is given as $\frac{N_B}{N_A} = 0.3$.Substituting the expressions: $\frac{N_0(1 - e^{-\lambda t})}{N_0 e^{-\lambda t}} = 0.3$.This simplifies to $\frac{1}{e^{-\lambda t}} - 1 = 0.3$,which means $e^{\lambda t} - 1 = 0.3$.So,$e^{\lambda t} = 1.3$.Taking the natural logarithm on both sides: $\lambda t = \ln(1.3)$.Since the decay constant $\lambda = \frac{\ln 2}{T}$,we have $\frac{\ln 2}{T} \cdot t = \ln(1.3)$.Solving for $t$: $t = T \frac{\ln(1.3)}{\ln 2} = T \frac{\log(1.3)}{\log 2}$.

$A$ radioactive nucleus $A$ with a half-life $T$ decays into a nucleus $B$. At $t = 0$,there is no nucleus $B$. At some time $t$,the ratio of the number of $B$ to that of $A$ is $0.3$. Then,$t$ is given by:

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