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half-life equations

A 64-g sample of germanium-66 is left undistured for 12.5 hours. At the end of that period, only 2.0 g remain. What is the half-life of this material?

Quote:

a 64-g sample of germanium-66 is left undistured for 12.5 hours. At the end of that period, only 2.0 g remain. What is the half-life of this material?

$N(t) = N_0 2^{-(\frac {t}{t_{\frac 12}})$

where $t_{\frac 12}$ is the half-life

Half-life - Wikipedia, the free encyclopedia

$2=64\,\times\,2^{-(\frac {12.5\,hr}{t_{\frac 12}})$

$\frac {1}{32}=2^{-(\frac {12.5\,hr}{t_{\frac 12}})$

Take the $log_2$ of both sides. Note that this is an unusual base, but it simplifies the processing since $log_2(\frac {1}{32})=log_2(\frac {1}{2^5})=log_2(2^{-5})=-5$ . You could use any logarithm base, but this makes the algebra easier.

$log_2(\frac {1}{32})=-5=-\frac {12.5\,hr}{t_{\frac 12}$

$t_{\frac 12} = \frac {12.5\,hr}{5}=2.5\,hr$

$5=\frac {12.5\,hr}{t_{\frac 12}$

$t_{\frac 12}=\frac {12.5\,hr}{5}=2.5\,hr$

there is one more way
calculate the number of atoms in 64g of germanium .This is u'r N0(pronounced as N not i.e initial no of atoms) Now calculate the number of germanium atoms in 2g Nt.
Now use the formula k=2.303 * log(No/Nt) / t where t-12.5 hrs
now t1/2 = 0.693/k