### Video Transcript

A radioactive sample initially contains 2.40 times 10 to the negative two moles of a radioactive material, which has a half-life of 6.00 hours. Approximately how many moles of the radioactive material remain after 6.00 hours? Approximately how many moles of the radioactive material remain after 12.00 hours? Approximately how many moles of the radioactive material remain after 36.00 hours?

We’re told in this statement that we initially start off with 2.40 times 10 to the negative two moles of a radioactive sample; we’ll call that initial amount 𝐴 sub zero. We’re also told that the half-life of this material is 6.00 hours; we’ll call that amount 𝑡 sub one-half. We want to know the amount of the material remaining after 6.00 hours, 12.00 hours, and 36.00 hours; we’ll call those amounts 𝐴 sub 6.00, 𝐴 sub 12.00, and 𝐴 sub 36.00, respectively.

This exercise is about radioactive decay, and specifically it’s about radioactive decay based on a sample’s half-life, which is the time it takes for that sample to lose half of its mass due to radioactive decay.

The mathematical relationship describing this decay says that 𝐴 sub 𝑓, the final amount of some substance, equals the initial amount of that substance, 𝐴 sub zero, times one-half raised to the 𝑡 divided by 𝑡 sub one-half power, where 𝑡 sub one-half is the material’s half-life.

When we apply this relationship to our scenario, the amount of the radioactive material that remains after 6.00 hours equals 𝐴 sub zero times one-half raised to the 6.00 divided by 𝑡 one-half power. Now because the time we’re considering, 6.00 hours, is identical to the half-life of our substance, we can expect that 𝐴 sub 6.00 will be one-half our initial amount 𝐴 sub zero.

When we plug in for 𝐴 sub zero and 𝑡 sub one-half, we see that our exponent in this equation is equal to one, which means it reduces to 𝐴 sub zero times one-half, or 1.20 times 10 to the negative two moles. That’s how much of the sample remains after 6.00 hours.

Now let’s consider how much of the sample remains after 12.00 hours have elapsed. All that changes in our basic equation is the time: 6.00 becomes 12.00. When we again enter in our values for the initial amount 𝐴 sub zero and 𝑡 sub one-half, the half-life, we see that, in this instance, the exponent goes to two.

One-half times one-half is one-fourth, and one-fourth of our initial amount, 𝐴 sub zero, is 6.00 times 10 to the negative third moles. That’s how much of the initial substance remains after 12.00 hours.

Now we solve for the amount remaining after 36.00 hours. We change the time value and again plug in for 𝐴 sub zero and 𝑡 sub one-half. This time, as we look at the exponent, we see 36 divided by six is six, and one-half raised to the sixth power is equal to one 64th. Multiplying this by our initial amount, 𝐴 sub zero, we find it’s 3.75 times 10 to the negative fourth moles. That’s how much of the sample remains after 36.00 hours.