Question

The radioactive isotope 239Pu has a half-life of approximately 24100 years. After 2000 years, there are 5g of 239Pu.(1) What was the initial quantity? (Round your answer to three decimal places.) g Tries 0/99(2) How much of it remains after 20000years? (Round your answer to three decimal places.) g Tries 0/99

Answer 1

The initial quantity of 239Pu can be calculated using the formula for radioactive decay, and it is approximately 43.931g. After 20000 years, the quantity remaining can be found using the same formula, and it is approximately 3.285g.

Step-by-step explanation:

The initial quantity of 239Pu can be determined by using the concept of exponential decay. The formula for radioactive decay is given by:

N(t) = N₀ * (1/2)^(t/T)

where N(t) is the quantity remaining after time t, N₀ is the initial quantity, t is the time, and T is the half-life.

In this case, we are given that after 2000 years, there are 5g of 239Pu. Plugging these values into the formula:

5 = N₀ * (1/2)^(2000/24100)

Solving for N₀:

N₀ = 5 / (1/2)^(2000/24100)

Calculating this value:

N₀ ≈ 43.931g (rounded to three decimal places).

After 20000 years, we can use the same formula to find the quantity remaining:

N(20000) = N₀ * (1/2)^(20000/24100)

Plugging in the value of N₀ from the previous calculation:

N(20000) = 43.931 * (1/2)^(20000/24100)

Calculating this value:

N(20000) ≈ 3.285g (rounded to three decimal places).

Answer 2

Using the following formula:

`\begin{gathered} N(t)=N_o(0.5)^{(t)/(t_(1/2))} \\ where: \\ N(t)=Remaining_{\text{ }}quantity_{\text{ }}after_{\text{ }}time_{\text{ }}t \\ N_o=Initial_{\text{ }}quantity \\ t=time_{\text{ }}in_{\text{ }}years \\ t_(1/2)=half-life=24100 \end{gathered}`

(1)

`\begin{gathered} t=2000 \\ t_(1/2)=24100 \\ N(2000)=5g \\ so: \\ 5=N_o(0.5)^{(2000)/(24100)} \\ N_o=\frac{5}{(0.5)^{(2000)/(24100)}} \\ N_o\approx5.296 \end{gathered}`

(2)

Using the initial quantity calculated previously:

`\begin{gathered} t=20000 \\ N(20000)=5.296(0.5)^{(20000)/(24100)} \\ N(20000)=2.979 \end{gathered}`

Answers:

For (1): 5.296

For (2): 2.979