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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#