Tuesday

October 21, 2014

October 21, 2014

Posted by **Joy** on Tuesday, August 20, 2013 at 9:23am.

- Math -
**Eren Jaeger**, Tuesday, August 20, 2013 at 10:16amN = (No)e^(-kt)

where

No is the original amount

k is constant

t is time

substitute,

(1/8)*No = No * e^(-k*150000)

the No cancels:

1/8 = e^(-150000k)

ln (1/8) = -150000k

k = -150000/ ln(1/8)

after solving for k, substitute to this equation and get the value of t (half-life):

1/2 * No = No * e^(-kt)

- Math -
**Joy**, Tuesday, August 20, 2013 at 10:24amWhat does the math symbols such as ^ and * mean? I still don't understand exactly what your trying to solve for. Am I trying to figure out the half-life of radioactive waste? Am I trying to figure out how many thousand years it takes in a half-life of radioactive waste? What does the term in math mean when saying "half-life"? Is the 1/8 suppose to stay as a fraction or be turned into a decimal or percent?

- Math -
**Eren Jaeger**, Tuesday, August 20, 2013 at 11:24amthe symbol ^ means "raise to". the symbol * means "multiplied by" as in for example

3^2 = 3*3 = 9

Yes, we're looking for the half-life. Half-life is the time it takes for a substance to achieve 1/2 of its original amount when it undergoes radioactive decay. the formula used for radioactive decay is

N = (No)*e^(-k*t)

where

N = remaining amount after time, t

No = the original amount

k = some constant

t = time

since we don't know the k value (nothing is given), as well as the original amount (the No), express N in terms of No so we can cancel them on both sides. that's why the equation becomes

(1/8)*No = No * e^(-k*150000)

It is stated in the problem that after 150000 years (this is the variable t), the remaining amount will only be 1/8 of the original (the original is No, right?).

since the equation above contains No term, we can cancel them:

(1/8) = e^(-k*150000)

then get the natural log of both sides. That will cancel the e on the right side, leaving only the exponent:

ln (1/8) = ln (e^(-k*150000))

ln (1/8) = -150000k

k = -150000/ ln(1/8)

Get a calculator and solve the above equation to get k.

Now, after getting k, you need to substitute it to the original equation:

N = (No)*e^(-k*t)

The N here will be equal to (1/2)*No since half-life (remember the definition?). Again, No will be cancelled on both sides. The equation becomes:

1/2 * No = No * e^(-kt)

1/2 = e^(-k*t)

ln (1/2) = ln(e^(-k*t))

ln (1/2) = -k*t

k = (-ln(1/2)) / t

now you know the value of k, solve for t.

- Math -
**Eren Jaeger**, Tuesday, August 20, 2013 at 11:27ami mean,

t = (-ln(1/2)) / k

we're solving for t. the units are in years.

- Math -
**Steve**, Tuesday, August 20, 2013 at 12:34pmWow - that's a lot of work.

each half-life diminishes the amount by half.

1/8 = 1/2 * 1/2 * 1/2

So, 150,000 years is 3 half-lives

The half-life is thus 50,000 years.

**Answer this Question**

**Related Questions**

Chemistry - Which radioactive sample would contain the greatest remaining mass ...

Math - D.E.Q. - The half-life of a radioactive isotope is the amount of time it ...

AP Environmental Science (URGENT) - Strontium-90 is a radioactive waste product ...

calculus - The rate at which an amount of a radioactive substance decays is ...

Calc - Radioactive radium () has a half-life of 1599 years. What percent of a ...

Math - The half-life of radioactive strontium-90 is approximately 32 years. In ...

Math - The half-life of radioactive strontium-90 is approximately 28 years. In ...

Calculus - [Exponential growth & decay] The half-life of radioactive strontium-...

Calculus - The half-life of radioactive strontium-90 is approximately 31 years. ...

Calculus - The half-life of radioactive strontium-90 is approximately 31 years. ...