Calculating When to Replace Cobalt-60

[priscilla98]

Homework Statement

Cobalt-60 has a half-life of 5.26 year. The cobalt-60 in a radiotherapy unity must be replaced when its radioactivity falls to 75 percent of the original sample. If the original sample was purchased in November of 1996, when will it be necessary to replace the cobalt-60?

Homework Equations

ln ( No / Nt ) = kt

t 1/2 = .693 / k

The Attempt at a Solution

I know we can find k constant by using t 1/2 = .693 / k.

k = .693 / 5.26
k = .1317

And we can use the ln ( No / Nt ) = kt and for No = 75 and Nt = 25.

- Am I doing this right? Thanks

 

[MysticDude]

Well since you are having a decay, your k value should actually be -.1318.

When it's radioactivity falls to 75% of the original, you will have only 1/4 of the original.
So now we have to play around with the exponential decay formula to get t. I'll give you the start.

\frac{1}{4} = e^{-.1318*t}, hope this gives you a start! Remember that we are looking for t in years :D

 

[priscilla98]

Well since you are having a decay, your k value should actually be -.1318.

When it's radioactivity falls to 75% of the original, you will have only 1/4 of the original.
So now we have to play around with the exponential decay formula to get t. I'll give you the start.

\frac{1}{4} = e^{-.1318*t}, hope this gives you a start! Remember that we are looking for t in years :D

Right, out of 75 percent, only 25 percent would be left. But I'm a little confused about this problem. Okay, when you plug in e^(-.1318) = .8765. Okay wait I see the approach you're taking to solve this.

1/4 = (.8765) (t)
.25 = (.8765) (t)
.2852 = t

 

[MysticDude]

Does it seem logical that 75% of a substance will be gone in .2852 years when half-life is only 5.26 years? No right?
I see your mistake though.
I would have done:
\frac{ln(\frac{1}{4})}{-.1318} = t I would get to this by taking the natural log of both sides only living me with ln(\frac{1}{4}) = (-.1318)t then dividing both sides by -.1318.

Do you understand what I'm saying?

 

[priscilla98]

Does it seem logical that 75% of a substance will be gone in .2852 years when half-life is only 5.26 years? No right?
I see your mistake though.
I would have done:
\frac{ln(\frac{1}{4})}{-.1318} = t I would get to this by taking the natural log of both sides only living me with ln(\frac{1}{4}) = (-.1318)t then dividing both sides by -.1318.

Do you understand what I'm saying?

Okay, I'm pretty sure I understand. Okay, therefore, it would be 10.52 years.

 

[MysticDude]

Yes and now to answer the question, we would have to add 10.52 years to the year 1996, which is 2006.52. So in 2006.52, you would have 75% of the Cobalt-60 gone :D

 

[priscilla98]

Okay thanks a lot.

- Have a good weekend :)

 

[Borek]

The cobalt-60 in a radiotherapy unity must be replaced when its radioactivity falls to 75 percent of the original sample.

Falls TO 75%, not BY 75%.

 

[MysticDude]

Falls TO 75%, not BY 75%.

OH GOSH, I must have read "Falls 75%"...wow.

In other words, t should be 2.1827 years. Making it 1998.1827. Wow, makes me wonder why I can screw up some physics problems. Thanks Borek.

SORRY priscilla98 :(

 

[priscilla98]

OH GOSH, I must have read "Falls 75%"...wow.

In other words, t should be 2.1827 years. Making it 1998.1827. Wow, makes me wonder why I can screw up some physics problems. Thanks Borek.

SORRY priscilla98 :(

It's okay but how did you get 2.1827 years? Thanks a lot

 

[MysticDude]

Well k is still the same, the only thing I changed is what is on the left of the equation.
I originally had this:
<br /> \frac{ln(\frac{1}{4})}{-.1318} = t<br />
but now I just had this:
<br /> \frac{ln(\frac{3}{4})}{-.1318} = t<br />

That's all :D