Radioactivity (Lecture 2) PDF

**RADIOACTIVITY (Lecture2)\ ** A nuclear reaction is different from a chemical reaction. In a chemical reaction, atoms of the reactants combine by a rearrangement of extra-nuclear electrons but the nuclei of the atoms remain unchanged. In a nuclear reaction, on the\ other hand, it is the nucleus of...

**RADIOACTIVITY (Lecture2)\ ** A nuclear reaction is different from a chemical reaction. In a chemical reaction, atoms of the reactants combine by a rearrangement of extra-nuclear electrons but the nuclei of the atoms remain unchanged. In a nuclear reaction, on the\ other hand, it is the nucleus of the atom which is involved. The number of protons or neutrons in the nucleus changes to form a new element. **A study of the nuclear changes in atoms is termed Nuclear Chemistry.** A number of elements such as uranium and radium are unstable. Their atomic nucleus breaks of its own accord to form a smaller atomic nucleus of another element. The protons and neutrons in the unstable nucleus regroup to give the new nucleus. This causes the release of excess particles and energy from the original nucleus, which we call **radiation.** The elements whose atomic nucleus emits radiation are said to be **radioactive.** The spontaneous breaking down of the unstable atoms is termed **radioactive disintegration or radioactive decay. The disintegration or decay of unstable atoms accompanied by emission of radiation is called Radioactivity.** **TYPES OF RADIATIONS** The radioactive radiations are of three types. These were sorted out by Rutherford (1902) by passing them between two oppositely charged plates. The one bending towards the negative plate carried positive charge and were named **alpha (α) rays.** Those bending towards the positive plate and carrying negative charge were called **beta (β) rays.** The third type of radiation, being uncharged, passed straight through the electric field and were named **gamma (ϒ) rays.** α, β and ϒ-rays could be easily detected as they cause luminescence on the zinc sulphide screen placed in their path. **PROPERTIES OF RADIATIONS** Alpha (α), beta (β) and gamma (ϒ) rays differ from each other in nature and properties. There chief properties are : (*a*) Velocity; (*b*) Penetrating power; (*c*) Ionisation. **ALPHA RAYS** **(1) Nature.** They consist of streams of α-particles. **By measurement of their e/m, Rutherford showed that they have a mass of 4 a.m.u and charge of +2.** They are **helium nucleus** and may be represented as ~2~^4^α or ~2~^4^He. **(2) Velocity.** α-particles are ejected from radioactive nuclei with very high velocity, about one-tenth that of light. **(3) Penetrating power:** Because of their charge and relatively large size, α-particles have **very little power of penetration** through matter. They are stopped by a sheet of paper, 0.01 mm thick aluminium foil or a few centimetres of air. **(4) Ionisation.** They cause **intense ionisation** of a gas through which they pass. On account of their high velocity and attraction for electrons, α-particles break away electrons from gas molecules and convert them to positive ions. **BETA RAYS** **(1) Nature.** They are streams of β-particles emitted by the nucleus. From their deflection electric and magnetic fields, Becquerel showed that β-**particles are identical with electrons.** They have very small mass (1/1827 a.m.u) and charge of -1. A beta (β)- particle is symbolized as ~-1~^0^β or ^0^*~-1~e*. **(2) Velocity.** They travel about **10 times faster than** α**-particles.** Their velocity is about the same as of light. **(3) Penetrating power.** β-**Particles are 100 times more penetrating in comparison to** α- **particles.** This is so because they have higher velocity and negligible mass. β-particles can be stopped by about 1 cm thick sheet of aluminium or 1 m of air. **(4) Ionisation.** The ionisation produced by β-particles in a gas is about **one-hundredth of that of** α**-particles.** Though the velocity of β-particles is higher but the mass being smaller, their kinetic energy is much less than α-particles. Hence they are poor ionisers. **GAMMA RAYS** **(1) Nature.** Unlike α and β-rays, they do not consist of particles of matter. ϒ**-Rays are a form of electromagnetic radiation** of shorter wavelength than X-rays. They could be thought of as high-energy photons released by the nucleus during α- or β-emissions. They have no mass or charge and may be symbolized as ~0~^0^γ. **(2) Velocity.** Like all forms of electromagnetic radiation, ϒ-**rays travel with the velocity of light.** **(3) Ionising power:** Their ionising power is very weak in comparison to α and β-particles. A ϒ-photon displaces an electron of the gas molecule to yield a positive ion. Since the chances of photon-electron collisions are small, ϒ-rays are weak ionisers. **(4) Penetrating power.** Because of their high velocity and non-material nature, ϒ-**rays are most penetrating.** They cannot be stopped even by a 5 cm thick sheet of lead or several metres thick layer of concrete. Table 1.0: Comparison of the radioactive particles PROPERTIES α - particle β - particle ϒ -- rays ------------------- ---------------------------------- ------------------------ ----------------------------------------- Nature He -- nucleus ~2~^4^He Fast electron ~-1~^0^e Electromagnetic radiation Velocity One --tenth of velocity of light Velocity of light Velocity of light Penetrating power low moderate High Stopped by: Paper or 0.01mm thick Al sheet 1cm of Aluminium Several cm thick lead or concrete layer **DETECTION AND MEASUREMENT OF RADIOACTIVITY** The radioactive radiation can be detected and measured by a number of methods. The important ones used in modern practice are listed below. **(1) Cloud Chamber:** This technique is used for detecting radioactivity. The chamber contains air saturated with water vapour. When the piston is lowered suddenly, the gas expands and is supercooled. As an α or β-particle passes through the gas, ions are created along its path. These ions provide nuclei upon which droplets of water condense. The trail or cloud thus produced marks the track of the particle. Similarly, α- or β-particles form a trail of bubbles as they pass through liquid hydrogen. The **bubble chamber method** gives better photographs of the particle tracks. **(2) Ionisation Chamber:** This is the simplest device used to measure the strength of radiation. An ionisation chamber is fitted with two metal plates separated by air. When radiation passes through this chamber, it knocks electrons from gas molecules and positive ions are formed. The electrons migrate to the anode and positive ions to the cathode. Thus a small current passes between the plates. This current can be measured with an ammeter, and gives the strength of radiation that passes through the ionisation chamber. **(3) Geiger-Muller Counter:** This device is used for detecting and measuring the rate of emission of α- or β-particles. It consists of a cylindrical metal tube (cathode) and a central wire (anode). The tube is filled with argon gas at reduced pressure (0.1 atm). A potential difference of about 1000 volts is applied across the electrodes. When an α- or β-particle enters the tube through the mica window, it ionises the\ argon atoms along its path. The argon ions (Ar^+^) are drawn to the cathode and electrons to anode. Thus for a fraction of a second, a pulse of electrical current flows between the electrodes and completes the circuit around. **Each electrical pulse marks the entry of one** α**- or** β**-particle into the tube and is recorded in an\ automatic counter.** The number of such pulses registered by a radioactive material per minute, gives the intensity of its radioactivity. **(4) Scintillation Counter:** Rutherford used a spinthariscope for the detection and counting of α-particles. The radioactive substance mounted on the tip of the wire emitted α-particles. Each particle on striking the zinc sulphide screen produced a flash of light. These flashes of light (scintillations) could be seen through the eye-piece. With this device it was possible to count α-particles from 50 to 200 per second. **A modern scintillation counter** also works on the above principle and is widely used for the measurement of α- or β-particles. Instead of the zinc sulphide screen, a crystal of sodium iodide with a little thallium iodide is employed. The sample of the radioactive substance contained in a small vial, is placed in a 'well' cut into the crystal. The radiation from the sample hit the crystal wall and produce\ scintillations. **These fall on a photoelectric cell which produces a pulse of electric current for each flash of light.** This is recorded in a mechanical counter. Such a scintillation counter can measure radiation up to a million per second.\ **(5) Film Badges:** A film badge consists of a photographic film encased in a plastic holder. When exposed to radiation, they darken the grains of silver in photographic film. The film is developed and viewed under a powerful microscope.\ As α- or β-particles pass through the film, they leave a track of black particles. These particles can be counted. In this way the type of radiation and its intensity can be known. However, ϒ-radiation darken the photographic film uniformly. The amount of darkening tells the quantity of radiation. **A film badge is an important device to monitor the extent of exposure of persons working in the vicinity of radiation.** The badge-film is developed periodically to see if any significant dose of radiation has been absorbed by the wearer. **TYPES OF RADIOACTIVE DECAY:** According to the theory put forward by Rutherford and Soddy (1903), radioactivity is a nuclearproperty. The nucleus of a radioactive atom is unstable. It undergoes decay or disintegration by\ spontaneous emission of an α- or β-particle. This results in the change of proton-neutron composition of the nucleus to form a more stable nucleus. **The original nucleus is called the parent nucleus and the product is called the daughter nucleus.** α-**Decay\ **When a radioactive nucleus decays by the emission of an α-particle (α-**emission**) from the nucleus, the process is termed α-decay. An alpha particle has four units of atomic mass and two units of positive charge. If *Z* be the atomic number and *M* the atomic mass of the parent nucleus, the daughter nucleus will have;\ atomic mass = *M* -- 4 and atomic number = *Z* -- 2 **Thus an** α**-emission reduces the atomic mass by 4 and atomic number by 2.\ **For example, Radium decays by α-emission to form a new element Radon as in: β**-Decay\ **When a radioactive nucleus decays by β-particle emission (β-**emission**), it is called β-decay.\ A free β-particle or electron does not exist as such in the nucleus. It is produced by the conversion of a neutron to a proton at the moment of emission.\ Neutron p + e This results in the increase of one positive charge on the nucleus. The loss of a β-particle from the nucleus does not alter its atomic mass. For a parent nucleus with atomic mass *M* and atomic number *Z*, the daughter nucleus will have\ atomic mass = M *and* atomic number = *Z* + 1 **Thus a** β-**emission increases the atomic number by 1 with no change in atomic mass.** An example of β-decay is the conversion of lead-214 to bismuth-214, It is noteworthy that a β-emission results in the production of an isobar. Thus, and are isobaric as they have the same mass number 214 but different atomic numbers (82 and 83). **One** α**-emission and two** β**-emissions yield an isotope.** Let us consider the following series of changes. The parent element emits an α-particle and subsequently two β-particles, resulting in the formation of which is an isotope of the parent. Both the parent and the end-product have the same atomic number 84 but different mass numbers (218 and 214). They are said to be an isotopes. Example 1: How many α and β - particles are emitted in passing down from to Solution Let "a" be the number of α -- particles and b the number of β- particles Comparing the mass numbers: 232 = 208 + (4xa) + (bx0) 232 = 208 + 4a + 0 232 -- 208 = 4a 24 = 4a a = 6. Also, compare the atomic number: 90 = 82 + (2xa) + (b x -1) 90 = 82 + 2a --b 90 -82 = 2a --b 8 = (2x6) -- b 8 = 12 --b 8 -- 12 = -b -4 = -b b = 4 Thus the number of α- particles emitted = 6 and the number of β- particles emitted = 4. **Example 2:** is a β-emitter and is an α-emitter. What will be the atomic\ masses and atomic numbers of daughter elements of these radioactive elements ? Predict the position of daughter elements in the periodic table. Solution underdoes β-decay i.e Comparing the atomic masses, we have: 210 = 0 + b b = 210 Comparing the atomic numbers, we are going to get: 82 = -1 + a a = 82 + 1 a = 83 Thus the daughter element will have the same atomic mass **210** and its atomic number will be **83.** It will occupy **one position right** to the parent element. \(b) undergoes α-decay i.e Comparing the atomic masses, we get; 236 = 4 + *b* *b* = **236 -- 4** **b = 232** and comparing the atomic number, we get 88 = 2 + *a* *a* = **88 -- 2** **a = 86** Thus the daughter element will have atomic mass **232** and its atomic number will be **86.** It will occupy **two positions to the left** of the parent element. **RADIOACTIVE DISINTEGRATION SERIES** A radioactive element disintegrates by the emission of an α- or β- particle from the nucleus to form a new '*daughter element*'. This again disintegrates to give another '*daughter element*'. The process of disintegration and formation of a new element continues till a non radioactive stable element formed as the product. **The whole series of elements starting with the parent radioactive element to the stable end-product is called a Radioactive Disintegration Series.** Sometime, it is referred to as a **Radioactive Decay Series** or simply **Radioactive Series.** All the natural radioactive elements belong to one of the three series : (1) The Uranium Series (2) The Thorium Series (3) The Actinium Series **The Uranium Series:** It commences with the parent element uranium-238 and terminates with the stable element lead-206. It derives its name from uranium-238 which is the prominent member of the series and has the longest half-life. **The Thorium Series:** It begins with the parent element thorium 232 and ends with lead-208 which is stable. This series gets its name from the prominent member thorium-232. **The Actinium Series:** It starts with the radioactive element uranium-235. The end-product is the stable element lead-207. This series derives its name from the prominent member actinium-227. **RATE OF RADIOACTIVE DECAY:** The decay of a radioactive isotope takes place by disintegration of the atomic nucleus. It is not influenced by any external conditions. Therefore the rate of decay is characteristic of an isotope and depends only on the number of atoms present. If *N* be the number of undecayed atoms of an isotope present in a sample of the isotope, at time *t*,.........................................................................1.......................................................................2 Where means the rate of decrease in the number of radioactive atoms in the sample; and is the proportionality factor. This is known as the **decay constant** or **disintegration constant.** Putting *dt* = 1 in equation (1) we have..........................................................................3 **Thus decay constant may be defined as the proportion of atoms of an isotope decaying per second.** **UNITS OF RADIOACTIVITY:** The standard unit of radioactivity (*i.e.* rate of disintegration) is *Curie* (c). A curie is a quantity of radioactive material decaying at the same rate as 1 g of Radium (3.7 × 10^10^ dps). Rutherford is a more recent unit. **HALF-LIFE\ The half-life or half-life period (t~1/2~) of a radioactive isotope is the time required for one-half of the isotope to decay. Or, it may be defined as the time for the radioactivity of an isotope to be reduced to half of its original value**. Half-life period is characteristic of a radioactive element. For example, the\ half-life of radium is 1620 years. This means that 1g of radium will be reduced to 0.5 g in 1620 years and to 0.25 g in further 1620 years; and so on. Some other radioactive elements may have half-life of a fraction of a second and for others it may be millions of years. The unit of half-life period is time^--\ 1^ **THE ACTIVITY OF A RADIOACTIVE SUBSTANCE:** It is defined as the **rate of decay or the number of disintegrations per unit time.** The activity of\ a sample is denoted by *A*. It is given by the expression : The unit of activity is the **curie (Ci)** which is the rate of decay of 3.7 × 10^10^ disintegrations per second. The S.I unit of activity is **becquerel (Bq)** which is defined as one disintegration per second. The activity of a radioactive sample is usually determined experimentally with the help of a ***Geiger --muller counter*** **CALCULATIONS OF HALF-LIFE** From equation (1) we can write:....................................................................3 Where x is a constant N~o~ is the number of atom at time t=0, X=ln N~o~ Substituting the value of *X* in (3) **2.303log..............................................................4** **2.303log2** **........................................................................5** The value of decay constant can be found experimentally by finding the number of disintegrations per second with the help of Geiger Muller counter. Hence, half-life of the isotope concerned can be calculated by using equation 5. **Example 1.** Calculate the half-life of radium-226 if 1g of it emits 3.7 × 10^10^ alpha particles per second. **Solution** Rate of decay = Rate of emission of α-particles; Recall that; N = 3.7 x 10^10^ per second The number of atoms of radium present (*N*) in 1g of sample = From equation (5) stated earlier **,** Substituting the value of  and *N* in equation (1) above; **Example 2:** Calculate the disintegration constant of cobalt 60 if its half-life to produce Nickel -- 60 is 5.2 years. **Solution** Substituting the value of , we have: Hence, **Example 3**: The half-life period of radon is 3.825 days. Calculate the activity of\ radon. (atomic weight of radon = 222). **Solution** where *dN* is the number of atoms disintegrating per second, λ is the decay constant and *N* is the number of atoms in the sample of radon. **Calculation of *N* :** From expression (1) above Mass of 6.02 × 10^23^ atoms of radon = 222 g Mass of 1.7653 × 10^16^ atoms of radon= = 6.51 x 10^-6^ g By definition, the activity of radon is its mass in grams which gives 3.7 × 10^10^ disintegrations per second. Therefore **activity of radon** = **6.51 × 10^--\ 6^ g curie.** **CALCULATIONS OF SAMPLE LEFT AFTER TIME *T*** It follows from equation stated earlier that: Knowing the value of  , the ratio of *N*~0~/*N* can be calculated. If the amount of the sample present to start with is given, the amount left after lapse of time *t* can be calculated. **Example 4:** Cobalt-60 disintegrates to give nickel-60. Calculate the fraction and the percentage of the sample that remains after 15 years. The disintegration constant of cobalt-60 is 0.13 yr^-1^ **SOLUTION** = 0.847 The fraction remaining is the amount at time *t* divided by the initial amount. Hence the fraction remaining after 15 years is 0.14 or 14 per cent of that present originally. **Example 5:** How much time would it take for a sample of cobalt-60 to disintegrate to the extent that only 2.0 per cent remains ? The disintegration constant  is 0.13 yr^--\ 1^. **Solution** From the equation above: t = 30 years. **TUTORIAL QUESTIONS** 1. A bone taken from a garbage pile buried under a hill-side had ^14^C/^12^C ratio\ 0.477 times the ratio in a living plant or animal. What was the date when the animal was buried? **ANS: 6100 YRS** 2. The amount of carbon-14 in a piece of wood is found to be one-sixth of its amount in a fresh piece of wood. Calculate the age of old piece of wood. **ANS: 14818.5 YEARS.** 3. A radioactive isotope has half-life of 20 days. What is the amount of isotope left over after 40 days if the initial amount is 5 g? **ANS: 1.25g** 4. Calculate the time required for a radioactive sample to lose one-third of the atoms of its parent Isotope, if the half -- life is 33 min. **ANS: Time (t) = 19.31 min.** 5. The half-life of radioactive isotope is 47.2 sec. Calculate N/N~o~ left after one hour. **ANS: 1.12 x 10^-23^** **NUCLEAR REACTIONS:** In a chemical reaction there is merely a rearrangement of extra-nuclear electrons. The atomic nucleus remains intact. A nuclear reaction involves a change in the composition of the nucleus. The number of protons and neutrons in the nucleus is altered. The product is a new nucleus of another atom with a different atomic number and/or mass number. Thus, **A nuclear reaction is one which proceeds with a change in the composition of the nucleus so as to produce an atom of a new element.\ **The conversion of one element to another by a nuclear change is called **transmutation.** We have already considered the nuclear reactions of radioactive nuclei, producing new isotopes. Here we will consider such reactions caused by artificial means. **DIFFERENCES BETWEEN NUCLEAR REACTIONS AND CHEMICAL REACTIONS** --------------------------------------------------------- **Nuclear Reactions** --------------------------------------------------------- **1.** Proceed by redistribution of nuclear particles.\ **2.** One element may be converted into another.\ **3.** Often accompanied by release or absorp\ tion of enormous amount of energy.\ **4.** Rate of reaction is unaffected by external\ factors such as concentration, temperature,\ pressure and catalyst. --------------------------------------------------------- ---------------------------------------------------- **Chemical Reactions** ---------------------------------------------------- **1.** Proceed by the rearrangement of extra\ nuclear electrons.\ **2.** No new element can be produced.\ **3.** Accompanied by release or absorption of\ relatively small amount of energy.\ **4.** Rate of reaction is influenced by external\ factors ---------------------------------------------------- **NUCLEAR FISSION REACTIONS: In these reactions an atomic nucleus is broken or fissioned into two or more fragments.** This is accomplished by bombarding an atom by alpha particles ( ), neutrons ( ), protons ( ), deutrons ( ), etc. All the positively charged particles are accelerated to high kinetic energies by a device such as a *cyclotron*. This does not apply to neutrons which are electrically neutral. The projectile enters the nucleus and produces an unstable '**compound nucleus**'. It decomposes instantaneously to give the products. **NUCLEAR FUSION REACTIONS: These reactions take place by combination or fusion of two small nuclei into a larger nucleus.** At extremely high temperatures the kinetic energy of these nuclei overweighs the electrical repulsions between them. Thus they coalesce to give an unstable mass which decomposes to give a stable large nucleus and a small particle as proton, neutron, positron, etc. For example, Two hydrogen nuclei,( ) , fused to produce a deuterium nucleus (). **NUCLEAR EQUATIONS:** Similar to a chemical reaction, nuclear reactions can be represented by equations. **These equations involving the nuclei of the reactants and products are called nuclear equations.** The nuclear reactions occur by redistribution of protons and neutrons present in the reactants so as to form the products. Thus the total number of protons and neutrons in the reactants and products is the same. Obviously, **the sum of the mass numbers and atomic numbers on the two sides of the equation must be equal.** If the mass numbers and atomic numbers of all but one of the atoms or particles in a nuclear reaction are known, the unknown particle can be identified. **How to write a nuclear equation** 1\) Write the symbols of the nuclei and particles including the mass numbers (superscripts) and atomic numbers (subscripts) on the left (reactants) and right (products) of the arrow. \(2) Balance the equation so that the sum of the mass numbers and atomic numbers of the particles (including the unknown) on the two sides of the equation are equal. Thus find the atomic number and mass number of the unknown atom, if any. \(3) Then look at the periodic table and identify the unknown atom whose atomic number is disclosed by the balanced equation. **EXAMPLES OF NUCLEAR EQUATIONS** (*a*) **Disintegration of radium-236** by emission of an alpha particle ( ), **Mass Number; REACTANT = 236 PRODUCTS = 232 + 4 = 236** **Atomic Number REACTANT = 88 PRODUCTS = 86 + 2 = 88** *b*) **Disintegration of phosphorus-32** by emission of a beta particle ( ), **Mass Number; REACTANT = 32 PRODUCTS = 0 + 32 = 32** **Atomic Number REACTANT = 15 PRODUCTS = -1 + 16 = 15** (*c*) **Fission of Argon-40** by bombardment with a proton **Mass Number; REACTANT = 40 + 1=41 PRODUCTS = 40 + 1 =41** **Atomic Number REACTANT = 18 + 1 =19 PRODUCTS = 19 + 0 = 19** (*d*) **Fission of uranium-235** by absorption of a neutron () **Mass Number; REACTANT = 235+1=236 PRODUCTS = 141+92+3=236** **Atomic Number REACTANT = 92+0=92 PRODUCTS = 56+36+0=92.** **Exercise:** Complete the nuclear equation and identify the missing element. **ARTIFICIAL RADIOACTIVITY:** Many stable nuclei when bombarded with high speed particles produce unstable nuclei that are radioactive. The radioactivity produced in this manner by artificial means is known as **artificial radioactivity.** **ENERGY RELEASED IN NUCLEAR REACTIONS:** According to Albert Einstein, mass can be converted into energy and *vice versa.* His famous equation relating mass and energy is:..............................................(1) where *E* = energy ; *m* = mass and *c* = velocity of light. In nuclear reactions, a change in mass, ▲*m*, is accompanied by release of energy, ▲*E*. Thus equation (1) may be written as............................................(2) **NUCLEAR BINDING ENERGY** Atomic nucleus is made of protons and neutrons closely packed in a small volume. Although there exist intensive repulsive forces between the component protons, the nucleus is not split apart. This is so because the nucleons are bound to one another by very powerful forces. **The energy that binds the nucleons together in the nucleus is called the Nuclear binding energy.** When a nucleus is formed from individual protons and neutrons, there occurs a loss of mass (mass defect). According to Einstein's theory, it is this mass defect which is converted into binding energy. **Hence binding energy is the energy equivalent of the mass defect.** The various nuclei have different binding energies. Binding energy is a measure of the force that holds the nucleons together. Hence an energy equivalent to the binding energy is required to disrupt a nucleus into its constituent protons and neutrons. Since the nuclear energy is of an extremely high order, it is not easy to fission a nucleus. **Calculation of Binding Energy:** The binding energy of a nucleus can be calculated from its mass defect by using Einstein's equation: Example 1: What is the binding energy for nucleus if its mass defect is 0.08181 a.m.u **Solution** = 7.4 x 10^19^ J/mol Number of nuclei in one mole is 6.02 × 10^23^ (Avogadro's Law). ∴ Binding energy for nucleus may be expressed as: = 1.2 x 10^-4^ J