Radioactivity homework question

[roam]

Homework Statement

A nuclear power plant breeds 1mg of ²³⁹Pu per week. What activity, in Bq, does that create?

Homework Equations

$R_0 = \lambda N_0$ (initial activity of the sample)

$R = R_0 e^{-\lambda t}$ (exponential behavior of the decay rate)

$T_{1/2}=\frac{ln 2}{\lambda}$ (Half-life)

The Attempt at a Solution

(a) First I find the number of moles in 1mg of ²³⁹Pu:

$n= \frac{1\times10^{-3}}{239}=4.184 \times 10^{-6}$

Now, I think to find N₀ I have to times n by 6.02x10²³ nuclei/mol. Which gives us N₀=2.518x10¹⁸.

But how can I find the decay constant λ, if we are not given the half life?

I tried out the 1 week as the half life but I didn't get the correct answer:

$\lambda = \frac{0.693}{604800 \ s} = 1.1458 \times 10^{-6}$

$R_0 = \lambda N_0 = 2.88 \times 10^{12} \ Bq$

$R = R_0 e^{-\lambda t} = 1.44 \times 10^{12}$

So, what should I do? The correct answer must be: $2.3 \times 10^6 \ Bq$.

 

[CompuChip]

The half-life is usually something that you look up. Back in my time () we had a book of tables where you could find this, nowadays we have Wolfram Alpha.

 

[roam]

Right, I see. So there's no equation or anything you can use in this situation if you don't have a table?

 

[phyzguy]

The following is a good source for nuclear data, including half-lives. I've also zoomed into the page on Pu-239

http://atom.kaeri.re.kr/
http://atom.kaeri.re.kr/ton/nuc12.html

 

[asteriatic]

To find the decay constant, you can use the given information that the power plant breeds 1mg of 239Pu per week. This means that after one week, the initial activity (R_0) will decrease by half, giving a half-life of one week. Therefore, the decay constant can be calculated using the half-life equation, T_{1/2}=\frac{ln 2}{\lambda}. Plugging in the half-life of one week, we get λ = 0.693/604800 s = 1.1458 x 10^-6 s^-1.

Using this value for λ, we can now calculate the initial activity (R_0) using the equation R_0 = λN_0. As you correctly calculated, N_0 = 2.518 x 10^18 nuclei. Plugging in these values, we get R_0 = 2.88 x 10^12 Bq.

To find the activity after one week (R), we can use the equation R = R_0e^-λt, where t is the time in seconds. Since one week is equivalent to 604800 seconds, we can plug in these values and solve for R. This gives us R = 1.44 x 10^12 Bq.

Finally, to find the activity per week, we can simply multiply R by the number of weeks (1 week in this case). This gives us an activity of 1.44 x 10^12 Bq per week, which is equivalent to 2.3 x 10^6 Bq, the correct answer.