00:00 Good day everyone. My name is Mrs. Akinola. I will be teaching you physics. Today we will be looking at radioactivity.

00:21 Let's quickly look at the objective of today's lesson.

00:25 At the end of today's lesson, you should be able to define radioactivity. That's number

00:31 one, number two. You should be able to define radioactive elements, mention the types

00:37 of radiations emitted from radioactive elements. You should be able to list the properties

00:43 of each type of radiation. You should be able to define half-life and decay constants and

00:50 sketch the diagram of the cake off and solve problems of half-life. Now let's define radioactivity.

00:59 Radioactivity is the spontaneous disintegration or decay of nucleus of the atom of an element

01:07 during which it emits alpha, beta or gamma rays and any of the combination of auditory

01:15 and energy. Anytime radioactivity is taking place is either an alpha or beta or gamma rays

01:24 emitted or it can be a combination of alpha and beta or it can be a combination of auditory

01:30 radiations and with the emission of energy. Then we have radioactive elements. They are the

01:39 elements that disintegrate spontaneously and emit radiation from their nucleus. Example

01:47 of radioactive elements are uranium, radium, polonium, ionium, sodium and so on and so

01:55 forth. We have three types of radiations that can be emitted from radioactive elements

02:00 and these radiations are alpha particles, beta particles and gamma rays. Alpha particles

02:07 usually denoted by symbol alpha. Beta particles usually denoted by symbol beta and gamma

02:15 ray is usually denoted by symbol gamma. Now let's quickly look at the properties of alpha

02:22 particles. Number one, they are positively charged. Number two, they are helium nuclei.

02:29 That is anytime an alpha particle is emitted from a radioactive element. Then that means

02:35 an atom of helium has been emitted and helium nuclei usually represented by 4 to H e.

02:43 They are relatively I mass. They have little penetrating power, few centimeter in A. Then

02:50 they are slightly deflected in magnetic field. They have ionization energy. They ionize

02:57 it and they cause zinc sulfide to flourish. That is they cause fluorescence in zinc sulfide.

03:04 Then let's look at the properties of beta particles. Beta particles are negatively charged.

03:10 They are high energy electron. They are usually represented by 0 minus 1 e. So anytime a

03:19 beta particle is emitted from a radiative substance, then they are atomic number. We increase

03:25 by 1. Why the mass number will remain unchanged? Because the mass number is a 0. Beta particles

03:33 are relatively light. They have medium penetrating power in A and several millimeters in a aluminum.

03:42 They are strongly deflected in magnetic field. They do not cause fluorescence in zinc sulfide.

03:48 Then they travel at approximately speed of light. That's about 2.9 times 10.8 meters per

03:56 second. Properties of gamma ray. The electromagnetic waves are short wavelengths like lights. They are

04:05 electrical in neutral. That is they have no charge. They have negligible mass. They have

04:10 high penetrating power in A and solid and several meters in lead. They are not deflected by magnetic

04:18 field. This is because they have no charge. They have small ionization energy and they travel

04:25 at speed of light. This speed of light is 3.0 times 10 raise to power 8 meter per second.

04:40 So if you look at this diagram, we have the 3 radiations emitted from the radiative elements

04:47 or radiative substance. You have the alpha particle. This is the alpha particle. The alpha particle,

04:55 the beta particle and gamma rays. You can see that the beta particle is strongly deflected.

05:01 Alpha particle is slightly deflected and the gamma ray, if you look at the gamma ray, there is no

05:08 any deflection. Now I want to look at radiative decay on that which we are talking about,

05:18 alpha live and decay constants. Alpha live is the time taken for alpha of the atom initially

05:24 present in an element to decay or it is the time taken for alpha the quantity of the radiative

05:31 nucleus to disintegrate. Let's take for example if the alpha of an element is 10 seconds. So

05:38 that means after every 10 seconds the element will decay to alpha of its original value. If the

05:44 original value is represented by N or N nodes. So after 10 seconds, the element will decay to N

05:51 over 2. After another 10 seconds, that's 20 seconds, the element will decay to N over 4. That's

05:58 N over 2 divided by 2. So N over 2 times 1 over 2 and that will be N over 4. So after another 10

06:14 seconds, it will decay towards N over 8 and so on and so forth. Then we want to look at the

06:26 decay constants. It is defined as the instantaneous rate of decay per unit atom of substance and we can

06:34 also write it as lambda. Lambda is the decay constant because to number of atoms disintegrating

06:45 per second over number of atoms presents initially presents a dissociation. So we can also

07:15 define the decay constant as the ratio of the number of atoms disintegrating per second to the

07:20 number of atoms presents at the beginning. So the rate of decay does the n dt is proportional to

07:27 the number of atoms originally presents. So if the number of atoms originally presents at time

07:34 t is given by N, then the number of disintegration per unit time can be given as follows. You said

07:44 that the rate of decay that's the rate of decay, the n that's dt minus the n dt is proportional

07:57 to N is directly proportional to N. The negative sign means that the number of atoms that is decaying

08:06 per unit time decreases. That is the number of atom decreases which time. So that's the reason

08:12 why we are having this negative sign. Then we can, if we want to remove this proportionality sign

08:17 to put an equality sign, you must multiply by a constant and that constant is lambda which is the

08:24 decay constant. So we can write that minus the n over dt equals to lambda n. So if you make

08:34 lambda the subject of formula you are dividing both side by n. So you have lambda equals to minus

08:42 1 over n the n over dt. Then it has also been established experimentally that the decay

08:55 constant is equal to 0.693 over tf. Tf is the f life. So we can write that, we can write

09:10 that lambda equals to 0.693 over tf. I said tf is the f life of the substance. So likewise,

09:31 if you cross multiply and make tf the subject of the formula, then we can also write that

09:36 tf equals to 0.693 over lambda. So the f life, the SI unit of f life is seconds, why

09:48 the SI unit of decay constant is a per second. So if you plot a graph of the number of atoms

09:55 against time, it will give you a decay curve. From the decay curve you can see that after every

10:00 f life, the atom is reducing by f. The atom of the elements are decaying into f of this

10:08 original value. So after the first f life, that's tf, you have n not over 2. The original

10:14 atom is n not. So after the first f life, the atom will decay to n not over 2. After the

10:21 second f life, that's 2tf. You see, it will decay to n not over 4. And after the third

10:28 f life, it decays to n not over 8. Now let's quickly look at this example.

10:38 A certain radioactive elements ask a f life of 10 years. How long will it take to lose 7

10:48 over 8 of these original atoms present? 2. How long it takes until 1 over 4 of the atoms

10:55 originally present remain unchanged. Now let's look at the first part of the question.

11:00 I said how long will it take to lose 7 over 8 of these original atoms present? So for these

11:07 elements to lose 7 over 8 of these original atoms present, that means 1 over 8 of the original

11:15 atoms most remain. So let's look at the solution. So if 7 over 8 of the atoms has been lost,

11:25 1 over 8 will remain. So let's n equals to the number of atoms originally present. So

11:32 after 10 years, because they said the f life of the element is 10 years. So after 10 years,

11:40 after 10 years, n over 2 atoms remain, atoms remain. The after another 10 years, that's after 20 years.

12:00 This after 20 years, n over 4 atoms remain. The after 30 years, n over 8 atoms

12:30 remain. So that means it will take 30 years to lose 7 over 8 of these original atoms. Because

12:42 if you subtract, we said n is the original atom. So if you subtract n over 8 from n, it will

12:49 give you 7 n over 8. So that means 7 over 8 of the original atom has been lost. Then the

12:57 second part, they said how long will it take for only one over 4 of the atoms to remain. So after 10

13:06 years, we said n over 2 atoms remains. After 20 years, n over 4 atoms remains. So it will take 20

13:14 years for only one over 4 of the atoms to remain unchanged. Example 2. The isotope of a nuclear

13:23 atom, as a half-life of 5.4 times 10 years to power 3 seconds. Calculate is the K constant. So from

13:31 this question, we are giving the, from the question we are giving the half-life. So half-life, that's

13:48 T of equals to 5.4 times 10 years to power 3 seconds. So we are asked to find the K constants. That's

14:02 lambda. Let's put question mark. And we know that lambda equals to 0.693 over T of. So we can also

14:15 substitute where we have T of. We put 5.4 times 10 years to power 3 seconds. So lambda will be

14:23 0.693 over 5.4 times 10 years to power 3 seconds. So if we have a 6, it will give you 1.283 times 10

14:43 years to power minus 4 per second. So we have come to the end of today's lesson. You'll be answering the

14:53 following question. Number one, define radioactivity. Number two, what do you understand by the time

15:01 radiative elements? Give two examples of radiative elements. Number three, mention three types of

15:08 radiations from radioactive elements. Number four, leave the properties of each type of radiation.

15:14 Number five, define the following terms, half-life, decay constants, sketch the decay curve. Then you

15:24 solve the problem, the last calculation under. Thank you for watching. My name is Mrs. Saquilla.

15:30 See you next time.