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Radioactive Isotope Isotope which is unstable. It emits radiation & changes into another kind of atom.

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Presentation on theme: "Radioactive Isotope Isotope which is unstable. It emits radiation & changes into another kind of atom."— Presentation transcript:

1 Radioactive Isotope Isotope which is unstable. It emits radiation & changes into another kind of atom.

2 Transmutation Reaction When one kind of nucleus changes into another kind of nucleus.

3 Stability Depends on neutron to proton ratio.

4 Which elements are unstable? All the elements with atomic number > 83 (or beyond Bismuth)

5 Types of Radiation Alpha, Beta, Gamma Separated by electric or magnetic fields.

6 Least penetration power Alpha radiation. Shielding can be paper or cloth.

7 Most penetration power Gamma radiation. Requires lead/concrete shielding.

8 2 He or 2  Symbol for alpha radiation 4 4

9 2 He or 2  Same as the nucleus of a helium atom 4 4

10 -1 e or -1  or  - or  Symbol for beta particle 00

11 -1 e or -1  or  - or  Fast moving electron originating from nucleus 00

12 +1 e or +1  or  + Symbol for positron. 00

13 0  or  Symbol for gamma radiation. 0

14 0 n or n Symbol for neutron 1

15 1 H or 1 p Symbol for proton 1 1

16 Decay Mode When radioactive isotopes emit radiation, the decay mode is the type of radiation they emit.

17 Alpha Decay Unstable nucleus emits an alpha particle. Atomic #  by 2. Mass #  by 4.

18 220 Fr  4  + 216 At Unstable nucleus emits an alpha particle. Atomic #  by 2. Mass #  by 4. 87 285

19 220 Fr  4  + 216 At Natural Transmutation. 1 term on reactant side. 87 285

20 220 Fr  4  + 216 At Balance nuclear equations using conservation of atomic number & conservation of mass number. 87 285 4 + 216=220 87= 2 + 85

21 27 Al + 4 He  1 n + X Use conservation of atomic number & conservation of mass number. 13 0 1 + Y=27 + 4 13 + 2 = 0 + Z 2 Y Z So Y = 30 So Z = 15 Now have 30 X. Use PT. 15 X is P

22 Artificial Transmutation Particle “bullet” hits target nucleus & new isotope is produced. 2 terms on reactant side.

23 Artificial Transmutation 32 S + 1 n  32 P + 1 H 16015 1 bullettarget

24 Artificial Transmutation Particle “bullet” may be proton or alpha particle. To react with a nucleus, must overcome + + repulsive forces by accelerating bullet to high speeds. Particle “bullet” may be a neutron. Neutrons have no charge, so no repulsive forces to overcome. No acceleration necessary.

25 Fission Fission is division. Large nucleus (U-235 or Pu-239) is split into 2 medium sized nuclei by a neutron bullet. Excess neutrons & a great deal of energy are also produced.

26 239 Pu + 1 n  90 Sr + 147 Ba + 3 1 n Fission 94003856

27 Fusion Fusion: U for unite and U for sun. Very small nuclei (H & He) are jammed together. Huge amounts of energy are released.

28 1 H + 2 H  3 He Fusion 1 1 2

29 Identify each of the rxns a) 1 n + 235 U  142 Ba + 91 Kr + 3 1 n + energy b) 59 Co + 1 n  60 Co c) 3 He + 1 H  4 He + 0 e d) 14 C  14 N + 0 e 09256360 270 2 1 2+1 67 fission Artificial transmutation fusion Natural transmutation

30 Half-Life Amount of time required for ½ of a radioactive sample to decay. Each isotope has its own half-life. Not affected by temperature, physical or chemical environment, or age of sample. The shorter the half-life, the more radioactive.

31 Length of Half-Life Source: http://www.dlt.ncssm.edu/TIGER/chem2.htm#nuclear

32 # of Half-Lives = Time elapsed Half-life

33 Half-Life Map Fraction Remaining # of Elapsed Half-lives Mass Remaining 1 ½ ¼ 1 / 8 1 / 16 1 / 32 1 / 64 0 1 2 3 4 5 6 This line depends on the information in the problem. M initial or M final  Divide by 2  Multiply by 2

34 After 62 hours, 1.0 g remains unchanged from a sample of K-42. How much K-42 was in the original sample? Look up HL of K-42 in table N: 12.4 h Calculate number of HL: 62 hours = 5 HL 12.4 h/HL

35 Half-Life Map Fraction Remaining # of Elapsed Half-lives Mass Remaining 1 ½ ¼ 1 / 8 1 / 16 1 / 32 1 / 64 0 1 2 3 4 5 6 M initial or M final  Divide by 2  Multiply by 2 1.0 g 2.04.08.016.032.0

36 Problem 2 80 milligrams of a radioactive substance decays to 10 milligrams in 30 minutes. Calculate the HL. NOTICE: They gave BOTH masses. 80 mg  40 mg  20 mg  10 mg Took 30 min or 3 HL to decay to 10 mg, so 1 HL = 10 min.

37 Uses of Radioisotopes 1.C-14 dating of organic material 2.U-238 dating of rocks 3.Chemical tracers 4.Medical: use isotopes with short half-lives that are quickly eliminated from body. (I-131, Co-60, Tc-99) 5.Irradiation of food & mail


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