Hi! can anyone help me on this Chemistry problem?

Question

The half-life for the radioactive decay of calcium-47 is 4.5 dd. If a sample has an activity of 3.0 CiCi after 9.0 dd , what was the initial activity of the sample? Express your answer using two significant figures.

Answer 1

The decay of calcium-47 is a first-order process, and we can use the following equation to relate the initial and final activities:

A = A0 e^(-kt),

where A0 is the initial activity, A is the final activity after time t, k is the rate constant, and e is the base of natural logarithms.

The half-life of calcium-47 is 4.5 days, which means that the rate constant is:

k = ln(2) / t1/2 k = ln(2) / 4.5 days k = 0.1543 d^-1

We are given that the final activity A is 3.0 Ci after 9.0 days. Substituting these values into the equation above, we get:

3.0 Ci = A0 e^(-0.1543 d^-1 * 9.0 d) 3.0 Ci = A0 e^(-1.3887) 3.0 Ci = A0 * 0.2496 A0 = 12.02 Ci

Therefore, the initial activity of the sample was approximately 12.02 Ci. Then for it to be rounded to two significant figures you would say 12 Ci