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

Answer:

$A = 2.2 \times 10^6 counts/second$

Explanation:

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

$A = A_o e^{-\lambda t}$

here we know that

$A_o = 3.7 \times 10^{10} counts/s$

also we know that

$\lambda = \frac{ln2}{5.271}$ 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

$A = (3.7 \times 10^{10})(e^{-\frac{ln2}{5.271}(74)})$

$A = 2.2 \times 10^6 counts/second$