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Answer 1 · 2017-08-11T09:36:57

The mass left after $12000y$ is $=1.81g$ and the time to reach $1g$ is $=16800y$

The half life is $t_(1/2)=5600y$

This means that half the amount present at time $t=0$ will decay after $5600y$

The radioactive constant is $lambda=ln2/t_(1/2)y^-1$

We can solve this problem with the equation

$m=m_0e^(-lambdat)$

The radioactive constant is $lambda=ln2/5600$

The initial mass is $m_0=8g$

Therefore, the mass left after $12000y$

$m=8*e^(-ln2*12000/5600)=8*e^(-2.14ln2)=1.81g$

Now,

The mass left is $m=1g$

We apply the same equation

$1=8*e^(-ln2/5600*t)$

$e^(-ln2/5600*t)=1/8$

$ln2/5600*t=ln8$

$t=ln8*5600/ln2=16800y$

The time is $=16800y$