Radioactivity of a sample decreases 90% after 10 years. What is the half-life of the sample?Note: Radioactivity follows first order kinetics.

Sure, let's solve this step by step:

1. First, we need to understand that the decay of radioactive substances follows an exponential decay law, which can be expressed as N = N0 * e^(-kt), where N is the final amount, N0 is the initial amount, k is the decay constant, and t is time.

2. In this case, we know that the radioactivity decreases by 90% after 10 years. This means that N/N0 = 0.1 (because only 10% remains).

3. We can substitute these values into the decay equation: 0.1 = e^(-k*10).

4. To solve for k, we can take the natural logarithm of both sides: ln(0.1) = -k*10.

5. Solving for k gives us k = -ln(0.1)/10.

6. The half-life (T) of a radioactive substance is related to the decay constant by the equation T = ln(2)/k.

7. Substituting the value of k we found, we get T = ln(2)/[-ln(0.1)/10].

8. Simplifying this gives us T = -10*ln(2)/ln(0.1).

9. Calculating this gives us a half-life of approximately 3.3 years.