Half-Life Calculation: Radioisotope

[UWMpanther]

[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.

 

[hage567]

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 $$\lambda$$), 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.

 

[rocomath]

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

$$t_{\frac {1}{2}} = \frac {\ln{2}}{k}$$

Take 3000 counts as $$A_{0}$$ and 2736 counts as $$A_{t}$$

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

*don't forget to convert your units.

 

[UWMpanther]

ok so $$\ln{\frac {[A]_{0}}{[A]_{t}}} = kt$$ is what I'm going to use to calculate k

and then i use $$t_{\frac {1}{2}} = \frac {\ln{2}}{k}$$ to calculate for $$t_{\frac {1}{2}}$$

 

[rocomath]

Did you get an answer?

 

[UWMpanther]

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

 

[CaptainZappo]

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

Looks good to me.