Answer :
The radioactive sample has 77 years
Step-by-step explanation:
The radioactive decay of a radioactive isotope is described by the following equation
[tex]m(t)=m_0 e^{-\lambda t}[/tex] (1)
where
[tex]m_0[/tex] is the initial mass of the sample
[tex]\lambda[/tex] is the decay constant
t is the time
The half-life of this sample is
[tex]t_{\frac{1}{2}}=13 y[/tex]
So the decay constant is given by
[tex]\lambda=\frac{ln(2)}{t_{1/2}}=\frac{ln(2)}{13}=0.053 y^{-1}[/tex]
We also know that the original sample had a mass of
[tex]m_0 = 770 g[/tex]
While the final sample has a mass of
[tex]m=13 g[/tex]
Therefore, we can now re-arrange eq.(1) to find the time t:
[tex]-\lambda t = ln(\frac{m}{m_0})\\t=-\frac{ln(\frac{m}{m_0})}{\lambda}=-\frac{ln(13/770))}{0.053}=77 y[/tex]
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