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**[SOLVED] Half Life Help**

## Homework Statement

The activity of a radioisotope is 3000 counts per minute at one time and 2736 counts per minute 48 hours later. What is the half-life of th radioisotope?

This is where I'm completely lost.

- Thread starter UWMpanther
- Start date

- #1

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The activity of a radioisotope is 3000 counts per minute at one time and 2736 counts per minute 48 hours later. What is the half-life of th radioisotope?

This is where I'm completely lost.

- #2

hage567

Homework Helper

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See here for some information:

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli2.html#c3

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli2.html#c2

You first need to figure out the decay constant (represented by [tex]\lambda[/tex]), which you can do by using the decay equation. Once you have that, you can find the half-life*. The equations you need are in the link. Give it a try and see what you come up with.

*Or you could just substitute the expression for lambda (which relates to the half-life) into the decay equation and solve for the half-life all in one go. Same thing.

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli2.html#c3

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli2.html#c2

You first need to figure out the decay constant (represented by [tex]\lambda[/tex]), which you can do by using the decay equation. Once you have that, you can find the half-life*. The equations you need are in the link. Give it a try and see what you come up with.

*Or you could just substitute the expression for lambda (which relates to the half-life) into the decay equation and solve for the half-life all in one go. Same thing.

Last edited:

- #3

- 1,752

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[tex]\ln{\frac {[A]_{0}}{[A]_{t}}} = kt[/tex]

[tex]t_{\frac {1}{2}} = \frac {\ln{2}}{k}[/tex]

Take 3000 counts as [tex]A_{0}[/tex] and 2736 counts as [tex]A_{t}[/tex]

Also, do you know how the half-life equation is derived? And what connects these 2 equations?

*don't forget to convert your units.

[tex]t_{\frac {1}{2}} = \frac {\ln{2}}{k}[/tex]

Take 3000 counts as [tex]A_{0}[/tex] and 2736 counts as [tex]A_{t}[/tex]

Also, do you know how the half-life equation is derived? And what connects these 2 equations?

*don't forget to convert your units.

Last edited:

- #4

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and then i use [tex]t_{\frac {1}{2}} = \frac {\ln{2}}{k}[/tex] to calculate for [tex]t_{\frac {1}{2}}[/tex]

- #5

- 1,752

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Did you get an answer?

- #6

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yeah I got 21661 mins which then I converted to hours and that is 361 hours.

- #7

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Looks good to me.yeah I got 21661 mins which then I converted to hours and that is 361 hours.

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