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

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(B) Let the initial number of nuclei of $A$ be $N_0$.At time $t$,the number of nuclei of $A$ is $N_A$ and $B$ is $N_B$.Given that $\frac{N_B}{N_A} = 0

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

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(B) Let the initial number of nuclei of $A$ be $N_0$.At time $t$,the number of nuclei of $A$ is $N_A$ and $B$ is $N_B$.Given that $\frac{N_B}{N_A} = 0.3$,so $N_B = 0.3 N_A$.The total number of nuclei remains constant: $N_0 = N_A + N_B = N_A + 0.3 N_A = 1.3 N_A$.Thus,$N_A = \frac{N_0}{1.3}$.Using the radioactive decay law,$N_A = N_0 e^{-\lambda t}$.Substituting $N_A$,we get $\frac{N_0}{1.3} = N_0 e^{-\lambda t}$,which implies $e^{\lambda t} = 1.3$.Taking the natural logarithm on both sides,$\lambda t = \ln(1.3)$.Since $\lambda = \frac{\ln 2}{T}$,we have $t = \frac{\ln(1.3)}{\lambda} = \frac{\ln(1.3)}{\ln 2} T$.Using the property of logarithms,$\frac{\ln(1.3)}{\ln 2} = \frac{\log 1.3}{\log 2}$.Therefore,$t = T \frac{\log 1.3}{\log 2}$.

$A$ 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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