PhysicsHalf-Life Problem

Joystar1977

Active member
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

Staff member
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 $k$.

Joystar1977

Active member
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

Staff member
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

Well-known member
MHB Math Helper
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.