The radioactive isotope thorium 234 has a half-life of approximately 578 hours. a. If a sample h

The radioactive isotope thorium 234 has a half-life of...

Key Concepts

Exponential Decay Model

The exponential decay model is expressed mathematically as m(t) = m? · e^(–?t) or equivalently as m(t) = m? / 2^(t/T?/?), where m? is the initial mass, ? is the decay constant, t is time, and T?/? is the half-life. This formula is widely used to predict remaining quantities at any given time.

Logarithmic Techniques in Decay Problems

Logarithmic techniques are used to solve for time or the decay constant in exponential decay equations. Taking logarithms of both sides of the decay equation allows one to isolate the exponent, making it possible to solve for the unknown variable when the decay reaches a given level.

Exponential Decay

Exponential decay is a process in which a quantity decreases at a rate proportional to its current value. This concept is fundamental in many fields, particularly in radioactive decay, where it models how unstable atoms lose energy over time.

Half-life

Half-life is the time required for a substance undergoing exponential decay to reduce to half of its initial quantity. It provides a measurable parameter to describe the decay rate and is essential for setting up the decay model in many scientific applications.

Transcript

00:01 All right.

00:01 So in this problem, we are going to be using the half -light formula, which is the following.

00:06 So we've got a equals a -not times one -half to the t over h.

00:15 And so what those letters all mean is a equals the final amount.

00:23 A -0 equals the initial amount.

00:32 T equals the time and h equals our half -life.

00:39 So we're going to be taking advantage of this formula throughout all the problems that we're going to do here today.

00:45 So in our first problem, we are looking to figure out an expression for the mass after a certain number of hours.

00:55 Now, we are given an initial amount, but we're not really given much else.

01:00 So we are going to put in the information that we were given in order to come up with our expression.

01:06 So we don't know our a, but we do know.

01:09 Know our a not, so we're going to put that right here.

01:12 And remember to always put your units times our one -half.

01:16 We do not know our t, but we were given our half -life value.

01:21 So we're going to put that right here.

01:23 So we're just plugging in the values that the problem actually gave us.

01:27 And that sums up all we have to do for part a.

01:31 Now we're going to go on to part b.

01:34 So part b, we are actually given a amount of time that they want us to put in, and then we want to find the final value.

01:42 So we actually have an expression all lined up.

01:44 So all we have to do is put in what they gave us.

01:47 So we have a equals our initial mass times our one half.

01:55 Now we have our t, which in this case is going to be 75 hours over 578.

02:04 So what we're going to do here is we're actually going to put this into the calculator and we should be good to go.

02:10 And so once we put it into the calculator, we get a certain value, which is approximately 58 .49.

02:20 And again, we want to put our units here, which in this case is milligrams.

02:25 Also, one thing that i would suggest to you do is figure out how many sig figs you want to include in this problem.

02:31 So i use two here.

02:32 If you wanted to use one, it could be 0 .5.

02:35 If you wanted to use more, you would extend that value and have more.

02:40 So just make sure that you know what kind of answer you're expected to give...

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