Question

The half-life of Radium-226 is 1590 years. If a sample contains 500 mg, how many mg will remain after 4000 years?

Answer 1

Answer:

87.43 mg

Step-by-step explanation:

Let the decay rate of Radium-226 is r% annually.

So, we can write the equation of decay as $x = P(1 - \frac{r}{100})^{t}$ ....... (1) where, x is final amount after decay of t years and P is initial amount of radium-226.

Hence, $\frac{P}{2} = P(1 - \frac{r}{100} )^{1590}$

⇒ $(1 - \frac{r}{100})^{1590} = 0.5$

⇒ $(1 - \frac{r}{100} ) = (0.5)^{\frac{1}{1590} } = 0.999564$

Now, we have to calculate the remaining amount of Radium-226 after 4000 years if the initial amount is 500 mg.

So, we can write $x = 500(1 - \frac{r}{100} )^{4000} = 500(0.999564)^{4000} = 87.43$ mg. (Answer)

Answer 2

Answer:

for the following variation of the problem

The half-life of radium-226 is about 1,590 years. How much of a 100mg sample will be left in 500 years? Write your answer rounded to the nearest tenth.

A≈80.41mg

Step-by-step explanation:

This problem requires two main steps. First, we must find the decay constant k. If we start with 100mg, at the half-life there will be 50mg remaining. We will use this information to find k. Then, we use that value of k to help us find the amount of sample that will be left in 500 years.

Identify the variables in the formula.

AA0ktA=50=100=?=1,590years=A0ekt

Substitute the values in the formula.

50=100ek⋅1,590

Solve for k. Divide each side by 100.

0.5=e1,590k

Take the natural log of each side.

ln0.5=lne1,590k

Use the power property.

ln0.5=1,590klne

Simplify.

ln0.5=1,590k

Divide each side by 1,590.

ln0.51,590=k (exact answer)

We use this rate of growth to predict the amount that will be left in 500 years.

AA0ktA=?=100=ln0.51,590=500years=A0ekt

Substitute in the values.

A=100eln0.51,590⋅500

Evaluate.

In 500 years, there will be approximately 80.4mg remaining.