Question

You are using a Co source labeled as having the activity of 3.7 x 1010 counts/second in 1945. What is this source's current activity? (The half-lfe of 0Co is 5.271 years.)

Answer 1

`A = 2.2 * 10^6 counts/second`

Step-by-step explanation:

As we know that activity of a radioactive substance is given as

`A = A_o e^(-\lambda t)`

here we know that

`A_o = 3.7 * 10^(10) counts/s`

also we know that

`\lambda = (ln2)/(5.271)` per years

now since the activity is given in year 1945

so till today the total time is

`t = 2019 - 1945 = 74 years`

now from above formula

`A = (3.7 * 10^(10))(e^{-(ln2)/(5.271)(74)})`

`A = 2.2 * 10^6 counts/second`