##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# Half-life - Concept

###### Kendal Orenstein

###### Kendal Orenstein

**Rutger's University**

M.Ed., Columbia Teachers College

Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.

**Half-life** is the concept of time required for half of radioactive isotope s nuclei to decay. The amount remaining is calculated as the *(initial amount) (1/2) (# of 1/2 lives)^n* in which the number of 1/2 lives is equal to the time elapsed over the length of half-life.

It might be wondering if some, is there's any radioactive on this earth constantly spontaneously decaying how do we actually still have radioactive isotopes if our earth is billions millions of years old? Well actually they take a long time for things to decay doesn't happen instantly sometimes it does but sometimes they happens every millions of years such as uranium takes milli- the half life of uranium takes millions of years meaning that when the year started we had a certain amount and now we had after 4.5 I think billion years and million years it's half of the volume uranium is actually left on this earth so what do I mean by half life? So half life is the time required for half of the radioisotopes nuclei to decay.

So let's take strontium 90 for example, it's our radioactive isotope it's half life is 29 years so if I have originally if I have no time pass by, no year so no half life I have my in my position 10 grams of strontium 90 okay so one half life flows by and that's 29 years so 29 years later I [IB] my 10 grams that I originally had, half of it is left as the definition of half life states so that means I have 5 grams of strontium 90 left okay. After 2 half life go by 58 years, my oroginal sample of 10 grams has gone through 2 half life. Half of it's gone after the first 29 years and again another half after the second 29 years give me 2.5 grams left after 58 years and so on and so forth until actually no atoms are left of strontium 90, so we could do this inside the chart and do this step by step by step by step and it'll take a long time or we can go in translate this into a formula. And the formula is basically says the amount remaining how much we have is equal to initial amount which [IB] which is what you did 10 times one half which is exactly what we did times to the nth power so the n is the number of half lifes we went we went through so first half I said one second life had one half times two which is squared open one half times one half which is squared so number of half life is n. Okay so we can actually breakdown that n too because we might not go we might want to know like fraction of half life though things that do not have half life so I'm expect that and even further and said the number of half life which is we're going to denote it n equals little t which is the time that went by divided by big T which is the length of the half life okay so let's actually put this and this problem together.

Alright so let's say we have iron 59 a radioactive isotope of iron is used in medicine to diagnose blood circulation disorders okay. The half life of iron 59 is 44 and a half days. How much of the of the 2 gram sample remain after 133.5 days. Okay so let's actually look at our formula and plug everything in so the amount remaining is what we're looking for so that's x, initially we had 2 grams where we're going to multiply by one half to the what power? We didn't tell us the number of half lifes we have to break it down from time elapsed which 44 and a half days oh no sorry actually 133 and a half days went by the half life was 44 and a half days, make sure when you're doing this that this, the time elapse of the time and this time the units of time for these two are the same so you can't have seconds on top and days in the bottom so make sure they're actually the same and when you [IB] when you 2 times one half to the 30, 133 over 5 this actually if you did it Math it actually equals 3 half lifes it's going to give me x equals 0.25, 0.25 grams of Fe59 leftover which makes sense it's less than half because it's more than one half life actually 3 half lifes went by so that actually should be a lot less than 2 grams so a lot of it decomposing and went off into change into something else.

Let's do something harder you might see a problem that's a little bit tougher and that's this so let's say we're talking about carbon 14 and carbon 14 is a radioactive isotope of carbon 12 I mean its actually [IB] atmosphere with atmosphere has plenty of it and it also it's always happening. There is a certain radio, ratio of this isotope in the atmosphere always so in this atmosphere it's actually used in, very helpful for plants to use for photosynthesis the plants then take this and use it for photosynthesis, animals then eat the plants and so now all of the living actually have some ratio of carbon 14 that's equal through whatever is in the atmosphere, so all living objects have this in their system and it's totally fine. What happens is over time it'll breakdown so we have carbon 14 and it will break down because it is radioactive and it will go through beta decay and it will change into nitrogen no big deal and this takes 5,730 years to occur okay so it's not going to this isn't actually happening in our lifetime but what happens when you die? This will start breaking down so we're not going to be adjusting that carbon 14 anymore and we're going to start it's going to stop breaking down and decomposing into this nitrogen.

So then if a fossil C-14 ratio is one sixteenth that of the atmosphere. How old is the fossil? So we're going to actually say the amount remaining is one sixteenth of the original ratios was, so we're going to say one sixteenth. The initial amount was 1 a ratio was the same as that of the atmosphere okay great times one half we want to figure out the number the time elapsed so we're looking for little t and we know the half life is 5,730 years. Now I'm going to go into logarithms I'm not going to explain logarithms if you want to see more on logarithms you might want to check out the Math videos but I'm just going to assume that you know logarithms and say because I'm looking for the exponent I'm going to take the natural log of base 5 so I'm going to say the ln of natural log of one sixteenth is equal to the natural log of 1 one half, because the natural log of 1 is 1. Natural log of one half times 2 to the 5730 and when natural logs with natural logs I can take this exponent and put it in front so it's now becomes, now it's not exponent anymore so I'm going to say okay the natural log of one sixteenth is equal to t, let me see, t over 5730 times the natural log of 2 so I'm sorry one half, I'm going to divide by the natural log of one half and so natural log of one sixteenth over the natural log of one half, I'm going to have to bring my calculator up to do that, this is 4 equals t over 5730 so then I'm going to just do basic algebra multiply 4 times 5730 and that gives me 22,920 years have passed by.

This is good this is called carbon dating and it's actually a good way to figure out how old a fossil is and this is how this ratio that they use and it's actually use half lifes, so half life is a is a great way to figure how old something is or how much you're going to have after a certain period of time.

Please enter your name.

Are you sure you want to delete this comment?

###### Kendal Orenstein

M.Ed., Columbia Teachers College

Her energy is contagious. She gets you excited about learning new concepts and makes it easy to understand. Her love for chemistry comes across in all her videos.

Youre like a genius chemistry guru--thanks for helping me get that A”

My professor is about as clear as mud - thanks for explaining this concept in 7 minutes - that i haven't gotten in 6 weeks of class. Awesome - thanks you!”

You are my go-to whenever I have a chem question. You explain everything so well. THANK YOU!!”

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete