Half-Life

[john3j]

In Calculus class we are talking about exponential and logarithmic functions and I came across this problem and do not know how to do this problem. Can anyone help me figure out a formula or how to complete this problem?

"The decay rate of a radioactive isotope is 6.5 percent per year. Find its half-life."

Thanks,
John

 

[HallsofIvy]

In Calculus class we are talking about exponential and logarithmic functions and I came across this problem and do not know how to do this problem. Can anyone help me figure out a formula or how to complete this problem?

"The decay rate of a radioactive isotope is 6.5 percent per year. Find its half-life."

Thanks,
John

So after one year, 100- 6.5= 93.5 percent will be left so if you start with C, you will have .935C left. After 2 years, 93.5% of that, \(\displaystyle 0.935(0.935C)= (.935)^2C\). How much will be left after 3 years? How much after t years? Do you see why this involves "exponetial and logarithmic functions"? What will t be when exactly (1/2)C is left?

 

[john3j]

So after one year, 100- 6.5= 93.5 percent will be left so if you start with C, you will have .935C left. After 2 years, 93.5% of that, \(\displaystyle 0.935(0.935C)= (.935)^2C\). How much will be left after 3 years? How much after t years? Do you see why this involves "exponetial and logarithmic functions"? What will t be when exactly (1/2)C is left?

In our book it gives us one example for half life, y=200e^-0.05t where y is the amount of the isotope in grams and t is the time in years. Could I solve using this same equation and just substitute 0.065 in for the 0.05?

 

[MarkFL]

In our book it gives us one example for half life, y=200e^-0.05t where y is the amount of the isotope in grams and t is the time in years. Could I solve using this same equation and just substitute 0.065 in for the 0.05?

No, that won't work since:

\(\displaystyle e^{0.065}\ne1-0.065\)

If you want to use that formula, you need to write:

\(\displaystyle A(t)=A_0e^{\ln(0.935)t}\)

This will simplify to the formula HallsofIvy is hinting at above (using the identities \(\displaystyle c\cdot\log_a(b)=\log_a(b^c)\) and \(\displaystyle a^{\log_a(b)}=b\)).

Once you have the formula, then set:

\(\displaystyle A(t)=\dfrac{1}{2}A_0\) and solve for \(\displaystyle t\).