The half-life of P-32 is 14.30 days. How many milligrams of a 20.00 mg sample of P-32 will remain after 85.80 days?

Question

19084 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-04T20:27:27

The amount remaining is $=0.3125mg$

enter image source here

The half life is $t_(1/2)=14.3d$

This means that after $14.3$ days $50%$ of the sample will remain

The time is $T=85.8$ days

This is equal to $=85.8/14.3=6$ half lives

Therefore,

The amount remaining is $=1/2^6*20=0.3125mg$

We can solve this problem with the equation

$m=m_0e^(-lambdat)$

The radioactive constant is $lambda=ln2/t_(1/2)=ln2/14.3$

Therefore,

$m=20*e^(-ln2*85.3/14.3)=20*e^(-6ln2)=20*e^-ln(2^6)=20*e^-ln64$

$m=20/64=0.3125mg$