Question

Question #dbed0

Answer 1

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

Explanation:

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`