Half-Life of Radioactive Waste: 150,000 Years

[Joystar77]

After 150 thousand years, only 1/8 of the original amount of a particular radioactive waste will remain. The half-life of this radioactive waste is how many thousand years?

150,000 years / 1/8
150,000 / .125
= 1,200,000

If this isn't correct is the 1/8 suppose to be stay as a fraction, turned into a decimal, or turned into a percent? Also, if this isn't correct can somebody explain to me what the term means saying "half-life"?

 

[MarkFL]

Re: Math Word Problem

Since less than half of the original sample is present, you should expect that the half-life is less than the time that has already elapsed.

A quick way to solve this particular problem is to observe that the original amount has been halved 3 times, since \(\displaystyle \frac{1}{8}=\left(\frac{1}{2} \right)^3\), and so we know the half-life is one-third of the elapsed time.

We won't always be presented with a current amount that is a power of one-half times the original amount, so a more general method would be to use:

\(\displaystyle A(t)=A_0\left(\frac{1}{2} \right)^{-\frac{t}{k}}=\frac{A_0}{2^{\frac{t}{k}}}\) where \(\displaystyle 0<k\in\mathbb{R}\) is the half-life.

Now, we are told:

\(\displaystyle A(150000)=\frac{A_0}{8}\)

and so we may state:

\(\displaystyle 2^{\frac{150000}{k}}=2^3\)

Now, equate exponents, and solve for $k$.

 

[Joystar77]

Re: Math Word Problem

Mark FL,

What shows up in my email as you working out this word problem shows something totally and completely different. I know obviously the problem that I worked out was wrong, but the question I asked was does the 1/8 stay as a fraction or be turned into a decimal or percent?

Here is what was sent to my email:
Since less than half of the original sample is present, you should expect that the half-life is less than the time that has already elapsed.

A quick way to solve this particular problem is to observe that the original amount has been halved 3 times, since \frac{1}{8}=\left(\frac{1}{2} \right)^3, and so we know the half-life is one-third of the elapsed time.

We won't always be presented with a current amount that is a power of one-half times the original amount, so a more general method would be to use:

A(t)=A_0\left(\frac{1}{2} \right)^{-\frac{t}{k}}=\frac{A_0}{2^{\frac{t}{k}}} where 0<k\in\mathbb{R} is the half-life.

Now, we are told:

A(150000)=\frac{A_0}{8}

and so we may state:

2^{\frac{150000}{k}}=2^3

Now, equate exponents, and solve for $k$.

My understanding of this math problem than what shows on this site seems to be a little different. Which is suppose to be the correct way?

 

[MarkFL]

What was sent to you by email is the text minus the MATH tags.

As you can see by this line:

\(\displaystyle A(150000)=\frac{A_0}{8}\)

I left the 1/8 as a fraction, since 8 is a power of 2, it made the last step easier. But suppose, instead we were told 1/10 of the original was left, then we would have:

\(\displaystyle 2^{\frac{150000}{k}}=10\)

I try to use rational numbers rather than decimal representations whenever possible. I just prefer that form.

Taking the natural logarithm of both sides, we would find:

\(\displaystyle \frac{150000}{k}\ln(2)=\ln(10)\)

Now, solving for $k$, we find:

\(\displaystyle k=\frac{150000\ln(2)}{\ln(10)}\approx45154.49934959717\)

Does it make sense to you that if the elapsed time is held constant, but the amount left is decreased, then the half-life decreases as well? A smaller half-life means the substance decays at a quicker rate.

 

[SuperSonic4]

After 150 thousand years, only 1/8 of the original amount of a particular radioactive waste will remain. The half-life of this radioactive waste is how many thousand years?

150,000 years / 1/8
150,000 / .125
= 1,200,000

If this isn't correct is the 1/8 suppose to be stay as a fraction, turned into a decimal, or turned into a percent? Also, if this isn't correct can somebody explain to me what the term means saying "half-life"?

half-life is the amount of time it takes for half a given substance to decay. It needs a large sample size so the law of large numbers can be used.

• At the start you have a fraction of 1 remaining (none has decayed)
  
• After one half-life you have 0.5 remaining
  
• After two half-lives you have 0.25 remaining

If you use the powers of two you have (where H/L is half-life for brevity's sake):

• Start = \(\displaystyle 2^0\)
• One H/L = \(\displaystyle 2^{-1}\)
• Two H/L = \(\displaystyle 2^{-2}\)

There is a pattern here: after n half-lives you have \(\displaystyle 2^{-n}\) left of the original amount.

If you know that \(\displaystyle \dfrac{1}{8} = 2^{-3}\) you can say that three half-lives have passed in those 150,000 years so one half-life must be one-third of the 150,000 years that have passed.

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More generally you can use the formula for exponential decay

\(\displaystyle A = A_0e^{-t / t_{1/2}}\) where:

• \(\displaystyle A\) is amount remaining at time \(\displaystyle t\)
• \(\displaystyle A_0\) is amount when \(\displaystyle t=0\)
• \(\displaystyle t\) is time
• \(\displaystyle t_{1/2}\) is the half-life

We know that:

• \(\displaystyle A = 0.125A_0\)
• \(\displaystyle A = A_0\)
• \(\displaystyle t = 150,000\)

You can then plug in those values and find \(\displaystyle t_{1/2}\)

\(\displaystyle 0.125A_0 = A_0 e^{-150000/t_{1/2}}\)

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A quick note on units: You can use any unit of time you like as long as your half-life and time share the same unit. Your value will come out in years if you apply the formula above.