Question

Nucleus A decays into the stable nucleus B with a half-life of 14.19 s. At t=0 s there are 1,891 A nuclei and no B nuclei. At what time will there be 832 B nuclei?

Answer 1

$N_{oA}=1891$

Using the decay formula:

$N=N_o(\frac{1}{2})^{\frac{T}{T_{1/2}}}$

Where:

$N=1891-832=1059$

So:

$\begin{gathered} 1059=1891(\frac{1}{2})^{\frac{T}{14.19}} \\ \frac{1059}{1891}=(\frac{1}{2})^{\frac{T}{14.19}} \\ 0.56=(\frac{1}{2})^{\frac{T}{14.19}} \\ \end{gathered}$

Solve for T:

$\begin{gathered} log(0.56)=\frac{T}{14.19}log(\frac{1}{2}) \\ T=14.19(\frac{log(0.56)}{log(\frac{1}{2})}) \\ T\approx11.87s \end{gathered}$

Answer:

11.87 seconds