# Question Video: Understanding Exponential Decay Mathematics • Higher Education

At the start of an experiment, a scientist has a sample that contains 250 milligrams of a radioactive isotope. The radioactive isotope decays exponentially at a rate of 1.3 percent per minute. a) Write the mass of the isotope in milligrams, 𝑚, as a function of the time in minutes, 𝑡, since the start of the experiment. b) Find the half-life of the isotope, giving your answer to the nearest minute.

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### Video Transcript

At the start of an experiment, a scientist has a sample which contains 250 milligrams of a radioactive isotope. The radioactive isotope decays exponentially at a rate of 1.3 percent per minute. a) Write the mass of the isotope in milligrams, 𝑚, as a function of the time in minutes, 𝑡, since the start of the experiment. And b) Find the half-life of the isotope, giving your answer to the nearest minute.

Remember, we can model exponential growth and decay using the formula 𝑃 of 𝑡 equals 𝑃 nought times 𝑒 to the power of 𝑘𝑡, where 𝑃 nought is the initial value of 𝑃, and 𝑘 is the rate of growth or decay. In our case, there are 250 milligrams of the sample at the start of the experiment. So, 𝑃 nought must be equal to 250. We said that 𝑘 is the rate of growth or decay. Since the isotope is decaying, this value is going to be negative. And 1.3 percent as a decimal is 0.013. So, 𝑘 is negative 0.013. We can therefore say that the function that describes the mass in terms of minutes is 𝑚 equals 250 times 𝑒 the power of negative 0.013𝑡.

We’re now going to look at part b). Here, it helps us to understand the definition of the phrase half-life. It’s the time taken for the radioactivity of the isotope to fall to half of its original value. In our case, that’s half of 250. That’s 125 milligrams. We need to work out for what value of 𝑡 𝑚 is equal to 125. We therefore set 𝑚 equal to 125 and solve for 𝑡.

We see that 125 equals 250 times 𝑒 to the power of negative 0.013𝑡. We then divide both sides of this equation by 250. 125 divided by 250 is 0.5. So, we see that 0.5 equals 𝑒 to the power of negative 0.013𝑡. We then take the natural logarithm of both sides of this equation to get the natural log of 0.5 equals the natural log of 𝑒 to the power of negative 0.013𝑡. But of course, the natural log of 𝑒 to the power of negative 0.013𝑡 is just negative 0.013𝑡. And so, our final step is to divide both sides of this equation by negative 0.013. That gives us 𝑡 equals 53.319 minutes which, correct to the nearest minute, is 53 minutes.

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