Question

Referred to the radioactive cobalt 60 which has a half-life of 5.3 years, an initial sample of cobalt 60 has an activity of 12,400 cpm. How much time has passed when the activity of the sample has dropped to 775 cpm?

Answer 1

Using the half-life of cobalt-60 (5.27 years), it has taken approximately 21.08 years for the activity of a cobalt-60 sample to decrease from 12,400 cpm to 775 cpm, indicating that 4 half-lives have passed.

Step-by-step explanation:

The decay of radioactive isotopes, like cobalt-60, is characterized by its half-life, which is the time required for half of the isotope to decay. Knowing the initial activity of a sample and its half-life, we can use the concept of half-lives to estimate the time passed for a decrease in activity. For cobalt-60, with a half-life of approximately 5.27 years, we can calculate the number of half-lives that have passed to go from an activity of 12,400 counts per minute (cpm) to 775 cpm.

To solve this problem, we use the formula for exponential decay:

A = A_0 (1/2)^(t/T)

where:

• A is the activity after time t,
• A_0 is the initial activity,
• t is the time elapsed, and
• T is the half-life of the substance.

Let's calculate the number of half-lives n that have passed using the following steps:

1. Divide the final activity by the initial activity: 775 cpm / 12,400 cpm = 0.0625.
2. Find the fraction of the remaining amount in terms of powers of 1/2: 0.0625 = (1/2)^4, which means 4 half-lives have passed.
3. Calculate the total time: time = n × half-life = 4 × 5.27 years.
4. Total time = 21.08 years.

Therefore, it has taken approximately 21.08 years for the activity of the cobalt-60 sample to drop from 12,400 cpm to 775 cpm.