0
$\begingroup$

Radioactive decay and other phenomena are measured in terms of half lives. I think this is just an easily-comparable way of stating a repeated percentage loss, but my mathematical knowledge is very weak.

As a crutch, I wrote a program to start with 1,000,000, multiply by 0.999, and repeat. I had it print a message every time it reached a halving - that is, when the remaining amount was 500,000 or less, then 250,000 or less, etc. I saw a consistent half life of 693 rounds:

reached 499900.2346477281  at round 693  - 693 rounds elapsed
reached 249900.24460085342 at round 1386 - 693 rounds elapsed
reached 124925.1909144911  at round 2079 - 693 rounds elapsed
reached 62450.132251566276 at round 2772 - 693 rounds elapsed
reached 31218.83576633967  at round 3465 - 693 rounds elapsed
reached 15606.30332502206  at round 4158 - 693 rounds elapsed
reached 7801.594694162155  at round 4851 - 693 rounds elapsed
reached 3900.019018238128  at round 5544 - 693 rounds elapsed
reached 1949.6204223478458 at round 6237 - 693 rounds elapsed
reached 974.6157066056886  at round 6930 - 693 rounds elapsed
reached 487.2106204235443  at round 7623 - 693 rounds elapsed
reached 243.55670347259502 at round 8316 - 693 rounds elapsed
reached 121.75405321597741 at round 9009 - 693 rounds elapsed
reached 60.864879771978956 at round 9702 - 693 rounds elapsed
...

Changing the decay percentage gives me a different half life, but still a consistent one.

This leads me to believe that any half life could be stated in terms of percentage decay, such as "X% of the amount decays every Y seconds/days/years". But all I have is an experiment, not a proof.

Is my idea mathematically correct?

$\endgroup$
1
$\begingroup$

There is indeed a consistent link between the two.

Here is a page which demonstrates how the half-life depends upon the rate constant:

https://proofwiki.org/wiki/Half-Life_of_Radioactive_Substance

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.