Half-Life, Nuclear Decay Related Problem (answer in units of Ci)

[Spartan Erik]

Homework Statement

The half-life of an isotope of phosphorus is 14 days. If sample contains 2.9 × 10^16 decays such
nuclei, determine its activity. Answer in units of Ci.

Homework Equations

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The Attempt at a Solution

I know that one Ci is equal to 3.7*10^10 bq (or decay per second), I'm just not sure how I can relate this to the half-life of phosphorus.

==EDIT==

Nevermind guys, I got the formula after some searching. Ultimately, (decays*ln2)/(half life*24*3600*3.7e10)

 

[anathema]

= 0.0012 Ci

Hello,

Thank you for your post. To determine the activity of the sample, we can use the following formula:

Activity = (decay rate) / (half-life)

First, we need to determine the decay rate of the sample. This can be done by dividing the number of decays by the half-life:

Decay rate = (2.9 × 10^16 decays) / (14 days) = 2.07 × 10^15 decays per day

Next, we need to convert this to the appropriate units of activity, which is curies (Ci). To do this, we can use the conversion factor you mentioned in your post:

1 Ci = 3.7 × 10^10 decays per second

Therefore, the activity of the sample can be calculated as:

Activity = (2.07 × 10^15 decays per day) / (3.7 × 10^10 decays per second) = 5.59 × 10^4 Ci

So, the activity of the sample is 5.59 × 10^4 Ci. I hope this helps! Let me know if you have any further questions.

 

[nlightner]

is the answer in Ci. Good job on finding the formula! Just to clarify, the formula you used is the activity equation, where the activity (in Ci) is equal to the number of decays multiplied by the natural logarithm of 2 (since half-life is related to decay by ln2) and divided by the product of the half-life (in seconds) and the conversion factor for Ci to Bq (3.7e10). So in this case, the activity would be (2.9e16*ln2)/(14*24*3600*3.7e10) = 1.11 Ci.