Question #228817

. Initially 100milligrams of a radioactive substance was present. After 6 hours the mass has decreased by 3%. The rate of decay is proportional to the amount of the substance present at time t. Determine the half-life of the radioactive substance.


1
Expert's answer
2021-08-24T07:54:07-0400



The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:


dQdt=rt\frac{dQ}{dt}=-rt


Which has the following solution:


Q(t)=Q(0)ertQ{(t)}=Q(0)e^{-rt}


In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.


After 6 hours the mass had decreased by 3%.

This means that . We use this to find r.


This means that Q(6)=(10.03)Q(0)=0.97Q(0)Q(6)=(1-0.03)Q(0)=0.97Q(0)


We use this to find r


Q(t)=Q(0)ertQ{(t)}=Q(0)e^{-rt}


0.97Q(0)=Q(0)e6r0.97Q(0)=Q(0)e^{-6r}


e6r=0.97e^{-6r}=0.97


ln(e6r)=ln(0.97)ln(e^{-6r})=ln(0.97)


6r=ln0.97-6r=ln0.97


r=ln0.976r=-\frac{ln0.97}{6}


r=0.0051r=0.0051


Q(t)=Q(0)e0.0051tQ{(t)}=Q(0)e^{-0.0051t}



Determine the half life of radio active substance.


This is t for which Q(t)=0.5Q(0)Q(t)=0.5Q(0). So


Q(t)=Q(0)e0.0051tQ(t)=Q(0)e^{-0.0051t}


0.5Q(0)=Q(0)e0.0051t0.5Q(0)=Q(0)e^{-0.0051t}


e0.0051t=0.5e^{-0.0051t}=0.5


ln(e0.0051t)=ln(0.5)ln(e^{-0.0051t})=ln(0.5)


0.0051t=ln0.5-0.0051t=ln0.5


t=0.50.0051{t}=\frac{0.5}{0.0051}


t=135.9t=135.9


The half-life of the radioactive substance is 135.9 hours



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS