Half Life Chemistry Problems?

Question

If you can please answer as many as you can with some "work."
I will be using these as a "Study Guide" for my Final Exam. Thank you!!

12. An isotope of cesium (cesium-137) has a half-life of 30 years. If 1.0 mg of cesium-137 disintegrates over a period of 90 years, how many mg of cesium-137 would remain?

13. A 2.5 gram sample of an isotope of strontium-90 was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific Test Site. The half-life of strontium-90 is 28 years. In what year will only 0.625 grams of this strontium-90 remain?

14. Actinium-226 has a half-life of 29 hours. if 100 mg of actinium-226 disintegrates over a period of 58 hours, how many mg of actinium-226 will remain?

Answer 1

12) 90 / 30 = 3 (this is the number of half-lives)



(/12)^3 = 0.125 (this is the decimal amount remaining)



1 mg times 0.125 = 0.125 mg



13)



(1/2)^n = 0.625



n log 0.5 = log 0.625



n = 0.678 (this is the number of half-lives that has elapsed)



28 x 0.678 = 18.984



That's 19 years, so 1979.



14) 58 / 29 = 2 half lives



100 ---> 50 ---> 25



25 mg



More of this? Why sure:



http://www.chemteam.info/Radioactivity/WS-Radioactivity3.html

Answer 2

This Site Might Help You.



RE:

Half Life Chemistry Problems?

If you can please answer as many as you can with some "work."

I will be using these as a "Study Guide" for my Final Exam. Thank you!!



12. An isotope of cesium (cesium-137) has a half-life of 30 years. If 1.0 mg of cesium-137 disintegrates over a period of 90 years, how many...

Answer 3

12) half life of 30 years.



1mg (30years) = 0.5mg

another 30 years = 0.25mg

another 30 years = 0.125mg



so 0.125mg remaining.





13) 2.5g half life 28 years.



2.5g (28years) = 1.25g

another 28years = 0.625g



So it will be 56years after the explosion so 2016





14)100mg and 29 hrs = 50mg

another 29hrs = 25mg



So you have 25mg remaining.