How to Calculate Initial Radioactive Activity for Patient Treatment?

[soul5]

Homework Statement

Ok so I'm doing this practise problem but I have no clue what to do.

"A radioactive sample intended for irradiation of a hospital patient is prepared
at a nearby lab. The sample has a half-life of 83.61 h. What should it's initial activity be if it's activity is to be 7.0 X 10 ^ 8 Bq when it is used to irradiate the patient for 24 h?

Homework Equations

T 1/2 = Ln2/λ

N = N0e^λt

The Attempt at a Solution

Would you just do this?


N = 7.0 X 10 ^ 8 Bq e ^(83.61 - 24)

?

please help

 

[phyzmatix]

Would you just do this?


N = 7.0 X 10 ^ 8 Bq e ^(83.61 - 24)

?

please help

Unfortunately not, start with

T1/2 = ln2/lambda
lambda = ln2/(T1/2)

and try again. Hope this helps

 

[nlightner]

I can provide some guidance on how to approach this problem. First, it is important to understand the concept of half-life in radioactive decay. Half-life is the amount of time it takes for half of a radioactive sample to decay into a stable form. In this case, the half-life of the sample is 83.61 hours.

To solve this problem, we can use the equation N = N0e^-λt, where N is the final activity, N0 is the initial activity, λ is the decay constant, and t is the time. We know that the final activity is 7.0 X 10^8 Bq and the time is 24 hours. We also know that the half-life is 83.61 hours.

To find the initial activity, we can rearrange the equation to solve for N0:

N0 = N/e^-λt

Substituting the given values, we get:

N0 = 7.0 X 10^8 Bq/e^(-ln2/83.61 h * 24 h)

Simplifying, we get:

N0 = 7.0 X 10^8 Bq/e^-0.033

N0 = 7.0 X 10^8 Bq/0.967

N0 = 7.24 X 10^8 Bq

Therefore, the initial activity of the sample should be 7.24 X 10^8 Bq in order to have an activity of 7.0 X 10^8 Bq after 24 hours of irradiation. I hope this helps you understand the problem better and how to approach similar problems in the future.