Question

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:

• A. $t=\frac{T}{\log \left(1.3\right)}$
• B. $t=\frac{T}{\log \left(1.3\right)}$
• C. $t=T\frac{\log 1.3}{\log 2}$
• D. $t=T\log \left(1.3\right)$

Answer 1

The correct option is C $t=T\frac{\log 1.3}{\log 2}$

$A\stackrel{T}{\to }B$
$\lambda =\frac{\ln 2}{T}$
$N_A=N_0{e}^{-\lambda t}$
$N_B=N_0(1-e^{-\lambda t})$
$\frac{N_B}{N_A}=0.3$
$\frac{N_0(1-e^{-\lambda t})}{N_0{e}^{-\lambda t}}=0.3$
$1-e^{-\lambda t}=0.3e^{-\lambda t}$
$1=1.3e^{-\lambda t}$
$\frac{1}{1.3}=e^{-\lambda t}$
$\ln \frac{1}{1.3}=-\lambda t=-\frac{\ln 2}{T}t$
$t=\frac{T\log 1.3}{\log 2}$