JK-TK/Newdata huggingface dataset:
iv) have high ionizing power thus cause a lot of damage to living cells. v) a nuclide undergoing α-decay has its mass number reduced by 4 and its atomic number reduced by 2 Examples of alpha decay 210 84 Pb -> x 82 Pb + 42He 2+ 210 84 Pb -> 206 82 Pb + 42He 2+
3 226 88 Ra -> 222 y Rn + 42He 2+ 226 88 Ra -> 222 86 Rn + 42He 2+ x y U -> 23490 Th + 42He 2+ 238 92 U -> 23490 Th + 42He 2+ x y U -> 23088 Ra + 2 42He 2+ 238 92 U -> 23088 Ra + 2 42He 2+ 210 84 U -> xy W + 10 α 210 84 U -> 17064 W + 10 α 210 92U -> xy W + 6 α 210 92U -> 18680W + 6 α (ii)Beta (β) particle decay The Beta (β) particle has the following main characteristic: i)is negatively charged(like electrons) ii)has no mass number and atomic number negative one(-1) therefore equal to a fast moving electron (0 -1e) iii) have medium penetrating power and thus can be stopped /blocked/shielded by a thin sheet of aluminium foil. iv) have medium ionizing power thus cause less damage to living cells than the α particle. v) a nuclide undergoing β -decay has its mass number remain the same and its atomic number increase by 1 Examples of beta (β) decay 1.23 x Na -> 2312Mg + 0 -1e 23 11 Na -> 2312Mg + 0 -1e 2. 234 x Th -> y91 Pa + 0 -1e 234 90 Th -> y91 Pa + 0 -1e 3. 20770Y -> x y Pb + 30 -1e 20770Y -> 207 73Pb + 30 -1e 4. x y C -> 147N + 0 -1e
C -> 147N + 0 -1e 5.
234 x Th -> y91 Pa + 0 -1e 234 90 Th -> y91 Pa + 0 -1e 3. 20770Y -> x y Pb + 30 -1e 20770Y -> 207 73Pb + 30 -1e 4. x y C -> 147N + 0 -1e
C -> 147N + 0 -1e 5. 1 x n -> y1H + 0 -1e 1 0 n -> 11H + 0 -1e 6. 42He -> 411H + x 0 -1e 42He -> 411H + 2 0 -1e 7. 22888Ra -> 22890Th + x β 22888Ra -> 22892Th + 4 β 8. 23290Th -> 21282Pb + 2 β + x α 23290Th -> 21282Pb + 2 β + 5 α 9. 23892U -> 22688 Ra + x β + 92U -> 22688 Ra + 2 β + 3 α 10. 21884Po -> 20682Pb + x β + 84Po -> 20682Pb + 4β + 3 α (iii)Gamma (y) particle decay The gamma (y) particle has the following main characteristic: i)is neither negatively charged(like electrons/beta) nor positively charged(like protons/alpha) therefore neutral. ii)has no mass number and atomic number therefore equal to electromagnetic waves. iii) have very high penetrating power and thus can be stopped /blocked/shielded by a thick block of lead.. iv) have very low ionizing power thus cause less damage to living cells unless on prolonged exposure.. v) a nuclide undergoing y -decay has its mass number and its atomic number remain the same. Examples of gamma (y) decay • 3717Cl -> 3717Cl + y • 146C -> 146C + y The sketch diagram below shows the penetrating power of the radiations from a radioactive nuclide.
iv) have very low ionizing power thus cause less damage to living cells unless on prolonged exposure.. v) a nuclide undergoing y -decay has its mass number and its atomic number remain the same. Examples of gamma (y) decay • 3717Cl -> 3717Cl + y • 146C -> 146C + y The sketch diagram below shows the penetrating power of the radiations from a radioactive nuclide. 5 radioactive nuclide sheet of paper aluminium foil thick block of lead (radiation source) (block α-rays) (block β-rays) block y-rays) α-rays β-rays y-rays The sketch diagram below illustrates the effect of electric /magnetic field on the three radiations from a radioactive nuclide Radioactive disintegration/decay naturally produces the stable 20682Pb nuclide /isotope of lead.Below is the 238 92 U natural decay series. Identify the particle emitted in each case
6 Write the nuclear equation for the disintegration from : (i)238 92 U to 23490 T 238 92 U -> 23490 T + 4 2 He 2+ 238 92 U -> 23490 T + α (ii)238 92 U to 222 84 Rn 238 92 U -> 22284 Rn + 4 4 2 He 2+ 238 92 U -> 22284 Rn + 4α 230 90 Th undergoes alpha decay to 222 86 Rn. Find the number of α particles emitted. Write the nuclear equation for the disintegration. Working 230 90 Th -> 222 86 Rn + x 4 2 He Method 1 Using mass numbers 230 = 222 + 4 x => 4 x = 230 - 222 = 8 x = 8 / 4 = 2 α Using atomic numbers 90 = 86 + 2 x => 2 x = 90 - 86 = 4 x = 4 / 2 = 2 α Nuclear equation 230 90 Th -> 222 86 Rn + 2 4 2 He
7 214 82 Pb undergoes beta decay to 214 84 Rn. Find the number of β particles emitted. Write the nuclear equation for the disintegration.
Working 238 92 U -> 206 82 Pb + x 0 -1 e + y 4 2 He Using Mass numbers only 238 = 206 + 4y => 4y = 238 - 206 = 32 y = 32 = 8 α 4 Using atomic numbers only and substituting the 8 α(above) 238 92 U -> 206 82 Pb + 8 4 2 He + x 0 -1 e 92 = 82 + 16 + - x => 92 – (82 + 16) = - x x = 6 β Nuclear equation 238 92 U -> 206 82 Pb + 6 0 -1 e + 8 4 2 He 298 92 U undergoes alpha and beta decay to 214 83 Bi. Find the number of α and β particles emitted. Write the nuclear equation for the disintegration. Working 298 92 U -> 210 83 Bi + x 4 2 He + y 0 -1 e Using Mass numbers only 298 = 214 + 4x => 4x = 298 - 214 = 84 y = 84 = 21 α 4 Using atomic numbers only and substituting the 21 α (above) 238 92 U -> 214 83Bi + 21 4 2 He + y 0 -1 e 92 = 83 + 42 + - y
8 => 92 – (83 + 42) = - x x = 33 β Nuclear equation 298 92 U -> 210 83 Bi + 21 4 2 He + 33 0 -1 e B:NUCLEAR FISSION AND NUCLEAR FUSION Radioactive disintegration/decay can be initiated in an industrial laboratory through two chemical methods: a) nuclear fission b) nuclear fusion. a)Nuclear fission Nuclear fission is the process which a fast moving neutron bombards /hits /knocks a heavy unstable nuclide releasing lighter nuclide, three daughter neutrons and a large quantity of energy. Nuclear fission is the basic chemistry behind nuclear bombs made in the nuclear reactors.
Working 298 92 U -> 210 83 Bi + x 4 2 He + y 0 -1 e Using Mass numbers only 298 = 214 + 4x => 4x = 298 - 214 = 84 y = 84 = 21 α 4 Using atomic numbers only and substituting the 21 α (above) 238 92 U -> 214 83Bi + 21 4 2 He + y 0 -1 e 92 = 83 + 42 + - y
8 => 92 – (83 + 42) = - x x = 33 β Nuclear equation 298 92 U -> 210 83 Bi + 21 4 2 He + 33 0 -1 e B:NUCLEAR FISSION AND NUCLEAR FUSION Radioactive disintegration/decay can be initiated in an industrial laboratory through two chemical methods: a) nuclear fission b) nuclear fusion. a)Nuclear fission Nuclear fission is the process which a fast moving neutron bombards /hits /knocks a heavy unstable nuclide releasing lighter nuclide, three daughter neutrons and a large quantity of energy. Nuclear fission is the basic chemistry behind nuclear bombs made in the nuclear reactors. The three daughter neutrons becomes again fast moving neutron bombarding / hitting /knocking a heavy unstable nuclide releasing lighter nuclides, three more daughter neutrons each and a larger quantity of energy setting of a chain reaction Examples of nuclear equations showing nuclear fission 10 n + 235 b U -> 9038 Sr + c 54Xe + 310 n + a 10 n + 2713 Al -> 2813 Al + y + a 10 n + 28a Al -> b11 Na + 42 He a0 n + 147 N -> 14b C + 10 n + 11 H -> 21 H + a 10 n + 235 92 U -> 95 42 Mo + 139 57 La + 210 n + 7 a b) Nuclear fusion Nuclear fusion is the process which smaller nuclides join together to form larger / heavier nuclides and releasing a large quantity of energy. Very high temperatures and pressure is required to overcome the repulsion between the atoms. Nuclear fusion is the basic chemistry behind solar/sun radiation.
The three daughter neutrons becomes again fast moving neutron bombarding / hitting /knocking a heavy unstable nuclide releasing lighter nuclides, three more daughter neutrons each and a larger quantity of energy setting of a chain reaction Examples of nuclear equations showing nuclear fission 10 n + 235 b U -> 9038 Sr + c 54Xe + 310 n + a 10 n + 2713 Al -> 2813 Al + y + a 10 n + 28a Al -> b11 Na + 42 He a0 n + 147 N -> 14b C + 10 n + 11 H -> 21 H + a 10 n + 235 92 U -> 95 42 Mo + 139 57 La + 210 n + 7 a b) Nuclear fusion Nuclear fusion is the process which smaller nuclides join together to form larger / heavier nuclides and releasing a large quantity of energy. Very high temperatures and pressure is required to overcome the repulsion between the atoms. Nuclear fusion is the basic chemistry behind solar/sun radiation. 9 Two daughter atoms/nuclides of Hydrogen fuse/join to form Helium atom/nuclide on the surface of the sun releasing large quantity of energy in form of heat and light. 21H + 21H -> abHe + 11H + a -> 32He 21H + 21H -> a + 11 H 4 11H -> 42He + a 147H + a -> 178O + 11 H C: HALF LIFE PERIOD (t1/2) The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount. It is usually denoted t 1/2. The rate of radioactive nuclide disintegration/decay is constant for each nuclide. The table below shows the half-life period of some elements. Element/Nuclide Half-life period(t 1/2 ) 238 92 U 4.5 x 10 6 C 88 Ra 15 P 84 Po 0.0002 seconds The less the half life the more unstable the nuclide /element. The half-life period is determined by using a Geiger-Muller counter (GM tube) .A GM tube is connected to ratemeter that records the count-rates per unit time.
The table below shows the half-life period of some elements. Element/Nuclide Half-life period(t 1/2 ) 238 92 U 4.5 x 10 6 C 88 Ra 15 P 84 Po 0.0002 seconds The less the half life the more unstable the nuclide /element. The half-life period is determined by using a Geiger-Muller counter (GM tube) .A GM tube is connected to ratemeter that records the count-rates per unit time. 10 This is the rate of decay/ disintegration of the nuclide. If the count-rates per unit time fall by half, then the time taken for this fall is the half-life period. Examples a)A radioactive substance gave a count of 240 counts per minute but after 6 hours the count rate were 30 counts per minute. Calculate the half-life period of the substance. If t 1/2 = x then 240 --x-->120 –x-->60 –x--->30 From 240 to 30 =3x =6 hours =>x = t 1/2 = ( 6 / 3 ) = 2 hours b) The count rate of a nuclide fell from 200 counts per second to 12.5 counts per second in 120 minutes. Calculate the half-life period of the nuclide. If t 1/2 =x then 200 --x-->100 –x-->50 –x--->25 –x--->12.5 From 200 to 12.5 =4x =120 minutes =>x = t 1/2 = ( 120 / 4 ) = 30 minutes c) After 6 hours the count rate of a nuclide fell from 240 counts per second to 15 counts per second on the GM tube. Calculate the half-life period of the nuclide. If t 1/2 = x then 240 --x-->120 –x-->60 –x--->30 –x--->15 From 240 to 15 =4x =6 hours =>x = t 1/2 = ( 6 / 4 )= 1.5 hours d) Calculate the mass of nitrogen-13 that remain from 2 grams after 6 halflifes if the half-life period of nitrogen-13 is 10 minutes.
e) What fraction of a gas remains after 1hour if its half-life period is 20 minutes? If t 1/2 = x then: then 60 /20 = 3x 1 --x--> 1/2 –2x--> 1/4 –3x---> 1/8
11 After the 3rd half-life 1/8 of the gas remain f) 348 grams of a nuclide A was reduced to 43.5 grams after 270days.Determine the half-life period of the nuclide. If t 1/2 = x then: 348 --x-->174 –2x-->87 –3x--->43.5 From 348 to 43.5=3x =270days =>x = t 1/2 = ( 270 / 3 ) = 90 days g) How old is an Egyptian Pharaoh in a tomb with 2grams of 14C if the normal 14C in a present tomb is 16grams.The half-life period of 14C is 5600years. If t 1/2 = x = 5600 years then: 16 --x-->8 –2x-->4 –3x--->2 3x = ( 3 x 5600 ) = 16800years h) 100 grams of a radioactive isotope was reduced 12.5 grams after 81days.Determine the half-life period of the isotope. If t 1/2 = x then: 100 --x-->50 –2x-->25 –3x--->12.5 From 100 to 12.5=3x =81days =>x = t 1/2 = ( 81 / 3 ) = 27 days A graph of activity against time is called decay curve. A decay curve can be used to determine the half-life period of an isotope since activity decrease at equal time interval to half the original
12 (i)From the graph show and determine the half-life period of the isotope. From the graph t 1/2 changes in activity from: ( 100 – 50 ) => ( 20 – 0 ) = 20 minutes ( 50 – 25 ) => ( 40 – 20 ) = 20 minutes Thus t ½ = 20 minutes (ii)Why does the graph tend to ‘O’?
If t 1/2 = x then: 100 --x-->50 –2x-->25 –3x--->12.5 From 100 to 12.5=3x =81days =>x = t 1/2 = ( 81 / 3 ) = 27 days A graph of activity against time is called decay curve. A decay curve can be used to determine the half-life period of an isotope since activity decrease at equal time interval to half the original
12 (i)From the graph show and determine the half-life period of the isotope. From the graph t 1/2 changes in activity from: ( 100 – 50 ) => ( 20 – 0 ) = 20 minutes ( 50 – 25 ) => ( 40 – 20 ) = 20 minutes Thus t ½ = 20 minutes (ii)Why does the graph tend to ‘O’? Smaller particle/s will disintegrate /decay to half its original. There can never be ‘O’/zero particles D: CHEMICAL vs NUCLEAR REACTIONS Nuclear and chemical reaction has the following similarities: (i)-both involve the subatomic particles; electrons, protons and neutrons in an atom (ii)-both involve the subatomic particles trying to make the atom more stable. (iii)-Some for of energy transfer/release/absorb from/to the environment take place. 13 Nuclear and chemical reaction has the following differences: (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom. Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element. Chemical reactions do not form new elements (iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy. Chemical reactions produce or absorb small quantity of heat/energy. (iv)Nuclear reactions are accompanied by a loss in mass/mass defect.Do not obey the law of conservation of matter. Chemical reactions are not accompanied by a loss in mass/ mass defect hence obey the law of conservation of matter. (v)The rate of decay/ disintegration of the nuclide is independent of physical conditions (temperature/pressure /purityp/article size) The rate of a chemical reaction is dependent on physical conditions (temperature/pressure/purity/particle size/ surface area) E: APPLICATION AND USES OF RADIOCTIVITY.
(iv)Nuclear reactions are accompanied by a loss in mass/mass defect.Do not obey the law of conservation of matter. Chemical reactions are not accompanied by a loss in mass/ mass defect hence obey the law of conservation of matter. (v)The rate of decay/ disintegration of the nuclide is independent of physical conditions (temperature/pressure /purityp/article size) The rate of a chemical reaction is dependent on physical conditions (temperature/pressure/purity/particle size/ surface area) E: APPLICATION AND USES OF RADIOCTIVITY. The following are some of the fields that apply and use radioisotopes; a)Medicine: -Treatment of cancer to kill malignant tumors through radiotherapy. -Sterilizing hospital /surgical instruments /equipments by exposing them to gamma radiation. b) Agriculture: If a plant or animal is fed with radioisotope, the metabolic processes of the plant/animal is better understood by tracing the route of the radioisotope. c) Food preservation: X-rays are used to kill bacteria in tinned food to last for a long time. d) Chemistry: To study mechanisms of a chemical reaction, one reactant is replaced in its structure by a radioisotope e.g. During esterification the ‘O’ joining the ester was discovered comes from the alkanol and not alkanoic acid. During photosynthesis the ‘O’ released was discovered comes from water. e) Dating rocks/fossils: The quantity of 14C in living things (plants/animals) is constant. 14 When they die the fixed mass of 14C is trapped in the cells and continues to decay/disintegrate. The half-life period of 14C is 5600 years . Comparing the mass of 14C in living and dead cells, the age of the dead can be determined. F: DANGERS OF RADIOCTIVITY. All rays emitted by radioactive isotopes have ionizing effect of changing the genetic make up of living cells. Exposure to theses radiations causes chromosomal and /or genetic mutation in living cells. Living things should therefore not be exposed for a long time to radioactive substances. One of the main uses of radioactive isotopes is in generation of large cheap electricity in nuclear reactors. Those who work in these reactors must wear protective devises made of thick glass or lead sheet.
Living things should therefore not be exposed for a long time to radioactive substances. One of the main uses of radioactive isotopes is in generation of large cheap electricity in nuclear reactors. Those who work in these reactors must wear protective devises made of thick glass or lead sheet. Accidental leakages of radiations usually occur In 1986 the Nuclear reactor at Chernobyl in Russia had a major explosion that emitted poisonous nuclear material that caused immediate environmental disaster In 2011, an earthquake in Japan caused a nuclear reactor to leak and release poisonous radioactive waste into the Indian Ocean. The immediate and long term effects of exposure to these poisonous radioactive waste on human being is of major concern to all environmentalists. G: SAMPLE REVISION QUESTIONS The figure below shows the behaviour of emissions by a radioactive isotope x. Use it to answer the question follow
15 (a) Explain why isotope X emits radiations. (1mk) -is unstable //has n/p ratio greater/less than one (b) Name the radiation labeled T (1mk) alpha particle (c) Arrange the radiations labeled P and T in the increasing order of ability to be deflected by an electric filed. (1mk) T -> P a) Calculate the mass and atomic numbers of element B formed after 21280 X has emitted three beta particles, one gamma ray and two alpha particles. Mass number = 212 – (0 beta+ o gamma + (2 x 4 ) alpha = 204 Atomic number = 80 – (-1 x3) beta + 0 gamma + (2 x 2 )) alpha =79 b)Write a balanced nuclear equations for the decay of 21280 X to B using the information in (a) above. 21280 X -> 20479B + 242He + 3 0-1 e + y Identify the type of radiation emitted from the following nuclear equations. (i) 146 C -> 147N + ……… β - Beta (ii) 11 H + 10 n -> 21H + …… y -gamma (iii) 23592 U -> 9542Mo + 13957La + 10 n +…… 7 β – seven beta particles
16 (iv) 23892 U -> 23490Th + … … α-alpha (v) 146 C + 11 H -> 157N + …… y-gamma X grams of a radioactive isotope takes 100 days to disintegrate to 20 grams.
Mass number = 212 – (0 beta+ o gamma + (2 x 4 ) alpha = 204 Atomic number = 80 – (-1 x3) beta + 0 gamma + (2 x 2 )) alpha =79 b)Write a balanced nuclear equations for the decay of 21280 X to B using the information in (a) above. 21280 X -> 20479B + 242He + 3 0-1 e + y Identify the type of radiation emitted from the following nuclear equations. (i) 146 C -> 147N + ……… β - Beta (ii) 11 H + 10 n -> 21H + …… y -gamma (iii) 23592 U -> 9542Mo + 13957La + 10 n +…… 7 β – seven beta particles
16 (iv) 23892 U -> 23490Th + … … α-alpha (v) 146 C + 11 H -> 157N + …… y-gamma X grams of a radioactive isotope takes 100 days to disintegrate to 20 grams. If the half-life period isotope is 25 days, calculate the initial mass X of the radio isotope. Number of half-lifes = ( 100 / 25 ) = 4 20g -----> 40g ----> 80g-----> 160g -----> 320g Original mass X = 320g Radium has a half-life of 1620 years. (i)What is half-life? The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years.
Number of half-lifes = ( 100 / 25 ) = 4 20g -----> 40g ----> 80g-----> 160g -----> 320g Original mass X = 320g Radium has a half-life of 1620 years. (i)What is half-life? The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years. Number of half-lifes = ( 3240 / 1620 ) = 2 1 mg ---1620---> 0.5mg ---1620----> 0.25mg If 1mg -> 2.68 x 1018 atoms Then 0.25 mg -> ( 0.25 x 2.68 x 1018 ) = 6.7 x 1017 Number of atoms remaining = 6.7 x 1017 Number of atoms disintegrated = (2.68 x 1018 - 6.7 x 1017 ) = 2.01 x 1018 The graph below shows the mass of a radioactive isotope plotted against time
17 Using the graph, determine the half – life of the isotope From graph 10 g to 5 g takes 8 days From graph 5 g to 2.5 g takes 16 – 8 = 8 days Calculate the mass of the isotope dacayed after 32 days Number of half lifes= 32/8 = 4 Original mass = 10g 10g—1st -->5g—2nd-->2.5—3rd –>1.25—4th -->0.625 g Mass remaining = 0.625 g Mass decayed after 32 days = 10g - 0.625 g = 9.375g A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214 83Bi (a)Write the nuclear equation for the radioactive decay 21286 X -> 214 83Bi + 242He + 0-1 e (b)What is the atomic number of X2? 86 (c) After 112 days, 1/16 of the mass of X2 remained.
The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years. Number of half-lifes = ( 3240 / 1620 ) = 2 1 mg ---1620---> 0.5mg ---1620----> 0.25mg If 1mg -> 2.68 x 1018 atoms Then 0.25 mg -> ( 0.25 x 2.68 x 1018 ) = 6.7 x 1017 Number of atoms remaining = 6.7 x 1017 Number of atoms disintegrated = (2.68 x 1018 - 6.7 x 1017 ) = 2.01 x 1018 The graph below shows the mass of a radioactive isotope plotted against time
17 Using the graph, determine the half – life of the isotope From graph 10 g to 5 g takes 8 days From graph 5 g to 2.5 g takes 16 – 8 = 8 days Calculate the mass of the isotope dacayed after 32 days Number of half lifes= 32/8 = 4 Original mass = 10g 10g—1st -->5g—2nd-->2.5—3rd –>1.25—4th -->0.625 g Mass remaining = 0.625 g Mass decayed after 32 days = 10g - 0.625 g = 9.375g A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214 83Bi (a)Write the nuclear equation for the radioactive decay 21286 X -> 214 83Bi + 242He + 0-1 e (b)What is the atomic number of X2? 86 (c) After 112 days, 1/16 of the mass of X2 remained. Determine the half life of X2
18 1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16 Number of t 1 /2 in 112 days = 4 t 1 /2 = 112 = 28 days 4 1.Study the nuclear reaction given below and answer the questions that follow.
Determine the half life of X2
18 1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16 Number of t 1 /2 in 112 days = 4 t 1 /2 = 112 = 28 days 4 1.Study the nuclear reaction given below and answer the questions that follow. 126 C --step 1-->127 N --step 2--> 1211Na (a)126 C and 146 C are isotopes. What does the term isotope mean? Atoms of the same element with different mass number /number of neutrons. (b)Write an equation for the nuclear reaction in step II 127 N -> 1211Na + 0 -1e (c)Give one use of 146 C Dating rocks/fossils: Study of metabolic pathways/mechanisms on plants/animals Study the graph of a radioactive decay series for isotope H below. (a) Name the type of radiation emitted when isotope (i) H changes to isotope J. Alpha-Mass number decrease by 4 from 214 to 210(y-axis) atomic number decrease by 2 from 83 to 81(x-axis) (ii) J changes to isotope K Beta-Mass number remains 210(y-axis) atomic number increase by 1 from 81 to 82(x-axis). 19 (b) Write an equation for the nuclear reaction that occur when isotope (i)J changes to isotope L 21081 J -> 21084L + 3 0 -1e (i)H changes to isotope M 21483 H -> 20682M + 3 0 -1e + 2 4 2He Identify a pair of isotope of an element in the decay series K and M Have same atomic number 82 but different mass number K-210 and M-206 a)A radioactive substance emits three different particles. Identify the particle: (i)with the highest mass. Alpha/ α (ii) almost equal to an electron Beta/ β 1.a)State two differences between chemical and nuclear reactions(2mks) (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element.