Half-life problem

[djjones1957]

This problem is driving me crazy! Please help.

The half-life of an element is 4.3 x 10^9 yr. How long does it take a sample of the element to decay to 3/5 of its original mass? Use A=Ao(1/2)^(t/T) where Ao is the initial amount, T is the half-life, and t is the time. (Express results in scientific notation, rounded to the nearest hundredth.)

Thanks,
Diane

 

[wjm11]

The half-life of an element is 4.3 x 10^9 yr. How long does it take a sample of the element to decay to 3/5 of its original mass? Use A=Ao(1/2)^(t/T) where Ao is the initial amount, T is the half-life, and t is the time. (Express results in scientific notation, rounded to the nearest hundredth.)

The original mass is 100%, or just 1. When 3/5 are left, that means that 60% , .6, is left.

A=Ao(1/2)^(t/T)
.6 = (1)(.5)^(t/(4.3 x 10^9))

Start by taking the log of both sides. Consider that

log(a^n) = (n)log(a)

Can you handle it from here?

 

[djjones1957]

Thank you for trying to help. Apparently I need to go back and review 'logs'... I am actually in a Calculus class and this was a review. I know the answer is 3.17 x 10^9 yr, but I answered wrong... could you just show me the problem worked out? Thanks, Diane

 

[BigGlenntheHeavy]

\(\displaystyle y(t) \ = \ Ae^{kt}, \ standard \ generic \ formula \ for \ growth \ or \ decay.\)

\(\displaystyle Now, \ when \ the \ time \ = \ 0, \ we \ have \ 100\% \ or \ unity \ of \ the \ element.\)

\(\displaystyle Hence, \ y(0) \ = \ 1 \ = \ Ae^{k(0)}, \ implies \ A \ = \ 1, \ so \ y(t) \ now \ = \ e^{kt}.\)

\(\displaystyle Therefore, \ y(4.3E9) \ = \ \frac{1}{2} \ = \ e^{k(4.3E9)}, \ \implies \ k(4.3E9) \ = \ -ln(2).\)

\(\displaystyle Hence, \ k \ = \ ln(2^{\frac{-1}{4.3E9}})\)

\(\displaystyle Ergo \ y(t ) \ = \ e^{ln\big(2^{\frac{-t}{4.3E9}}\big)} \ = \ 2^{\frac{-t}{4.3E9}}\)

\(\displaystyle Now \ when \ y(t) \ decays \ to \ \frac{3}{5} \ = \ .6 \ of \ its \ original \ mass, \ we \ have:\)

\(\displaystyle .6 \ = \ 2^{\frac{-t}{4.3E9}}, \ \implies \ \frac{-t}{4.3E9} \ = \ \frac{ln(.6)}{ln(2)} \ \implies \ t \ = \ \frac{-4.3E9(ln(.6))}{ln(2)} \ \dot= \ 3.167E9 \ yrs.\)

 

[djjones1957]

Thank you Big Glenn!