Activity

Transcript

0:00Introduction

Hello everyone! Let me start with a puzzle. Put a counter near a radioactive source, and it clicks again and again. What exactly is it counting? It is counting the radiation from decays, nuclei breaking up one at a time. Now here is the strange part. The curie, a unit for this, equals three point seven times ten to the power ten decays per second. Why such an odd number? By the end of this video, you will calculate it yourself. Activity questions come up often in JEE, so let's begin.

0:35Activity: the rate of decay

So, the activity of a sample is its rate of decay. Capital N is the number of nuclei, and d N over d t is its rate of change. The two straight bars around it mean modulus, the size without the sign, and capital A here is the activity. Careful, in this chapter capital A is also the mass number, but here it means activity. Okay? Now tell me, why do we need the modulus bars? What is the sign of d N over d t? Yes, exactly, it is negative, because N keeps going down. The decay law says d N over d t equals minus lambda N, where lambda, the Greek letter, is the decay constant. The double arrow on screen means "so", and taking the size, the activity equals lambda times N, right? Now, from the decay law, N equals N naught, e to the power minus lambda t, where small t is the time. N naught, with the small zero below, is the value at time zero. So if the activity is lambda times N, how does it fall with time? It falls in exactly the same way, because lambda is a constant. So the activity equals A naught, e to the power minus lambda t, where A naught is the activity at time zero. Clear? And the activity falls with the same half-life as the number of nuclei, T half, written T with a small one half below. So after each half-life, the activity is halved. Got it?

2:08Units of activity

Now, the units. The first is the becquerel, written capital B, small q. One becquerel means one decay per second. Simple, isn't it? Next, the curie, written capital C, small i. One curie equals three point seven times ten to the power ten becquerel. That is thirty seven billion decays every second. With me? And this line holds the key to our puzzle. The wavy equals sign means approximately, so one curie is about the activity of one gram of radium; the small g is for gram. In example two, we will check this ourselves. Okay? One more unit is the rutherford, and one rutherford equals ten to the power six becquerel, that is, a million decays per second.

2:56Activity from the mass of a sample

Right? But problems usually give the mass of a sample, not the number of nuclei. So how do we find capital N from the mass? Yes, exactly, we go through the number of moles. Small m is the mass of the sample, and capital M is the molar mass. So m over M is the number of moles, and we multiply it by N with a small A below, which is Avogadro's number. Clear? Since the half-life is the natural log of two over lambda, lambda equals zero point six nine three over T half, because zero point six nine three is the natural log of two. Got it? Now put the two together. The activity equals zero point six nine three over T half, times m times N A, all over capital M, the molar mass. Okay? Now a trap: suppose the half-life is given in years, and you put it straight into this formula. Will your answer come out in becquerel? No, it will not, so be careful. One becquerel is one decay per second, so for the activity in becquerel, T half must be in seconds. Convert first, then substitute. Right?

4:05Trap: is the bigger sample more active?

Here is a quick idea that saves time. Putting in lambda, the activity equals zero point six nine three times N, over T half. The sign that looks like an open alpha means proportional to, so the activity is proportional to N over T half. With me? So, sample P has twice as many nuclei as sample Q, and its half-life is twice that of Q, okay? Which sample is more active? If you said P, because it has more nuclei, then careful, that is the trap. P has double the N, but also double the T half, and in N over T half the two twos cancel. The small P and Q below just name the samples. So the activities are equal. Clear?

4:49Example 1: 8 mCi source after three half-lives

Let's try an easy one first. The source has an activity of eight millicurie, written m C i, which is a thousandth of a curie. Its half-life is three hours, with the letter h for hours. Find its activity after nine hours. Pause the video and try it first. Done? Okay, let's check. First, how many half-lives fit in nine hours? Nine over three gives three, so small t, the time, is three half-lives. Right? The activity halves every half-life, remember? So it halves three times: eight becomes four, four becomes two, and two becomes one, and each arrow on screen is one halving. Got it? So the activity is eight over two cubed, which is one millicurie. See how quick that was? When the time is a whole number of half-lives, just keep halving.

5:42Example 2: activity of 1 g of radium-226 in becquerel

Clear? Now the harder one, and it settles our puzzle. Find the activity of one gram of radium two hundred and twenty six, whose half-life is one thousand six hundred years. On screen, R a is radium, and the small two two six at its top left is the mass number. Pause the video and try it first, and keep the half-life in seconds. Done? Okay, let's check. The molar mass is two hundred and twenty six grams per mole, so one gram is one over two hundred and twenty six of a mole. The number of nuclei is Avogadro's number, six point zero two times ten to the power twenty three, over two hundred and twenty six. That is about two point six six times ten to the power twenty one. With me? Next, the half-life in seconds. One year is about three point one six times ten to the power seven seconds, so we multiply one thousand six hundred by that; the small s on screen is for seconds. Okay? So lambda is zero point six nine three over that product, which gives one point three seven times ten to the power minus eleven per second; the s to the power minus one on screen is read as per second. Right? Now multiply lambda by N, the number of nuclei. What do you get, and have you seen that number before? Keeping all the digits, lambda times N comes to about three point six five times ten to the power ten, which rounds to three point seven times ten to the power ten becquerel. Got it? And that is one curie, the number from the start of this video! So the curie's strange number is roughly the activity of one gram of radium. That is our puzzle solved. See how it all fits together?

7:31Quick revision: A = lambda N, curie, activity from mass

So, quick revision. Activity is the rate of decay, lambda times N, and it falls as A naught, e to the power minus lambda t. One curie is three point seven times ten to the power ten becquerel, and the activity halves every half-life. And from the mass, use zero point six nine three over T half, times m times N A, over capital M, with T half in seconds. Now go and try the practice problems on this page. You can do them, I am sure. All the best!