Topic 2 Atomic Theory SL+HL

Topic 2.1 The atom Position ChargeRelative Mass Proton; p + Nucleus 1+ 1 Neutron; n Nucleus 0 1 Electron; e - Cloud/orbitals 1- 5*10 -4 around the ~ 0 nucleus Subatomic particles

An atom has no net charge. Atomic ions have lost or gained electrons and have a charge The number of Protons in an atom  The Element  Atomic number  Z  Number of Electrons

Isotopes Atoms with same number of Protons but different number of Neutrons are isotopes They are the same element = same Atomic number = Z Isotopes have different mass numbers Mass number = A = no. of protons + no. of neutrons The chemical properties is the same The physical properties can differ a little

The Symbol for Isotopes/Atoms

+ - Nucleus Electron H 1 Atomic number 1 Mass number Isotopes of hydrogen Natural abundance 99,99 %

+ - Electron H 1 2 Isotopes of hydrogen- Deuterium Nucleus Mass number Atomic number Heavy water- D 2 O natural abundance 0,01 %

+ - Electron H 1 3 Isotopes of hydrogen- Tritium Nucleus Mass number Atomic number Radioactive

- Alpha radiation + + Beta radiation Gamma radiation He -- e-e-  Radioactivity

Halflife Radiation intensity t ½ = 1600 år Ra Rn  Radium-226 Radon-222 alfaparticlegammaradiation He 2 4

Radioisotopes Not stable isotopes 14 Carbon C less than 0.001% of carbon, half life 5730 years. 14 C  14 N + e - 14 N + cosmic radiation  14 C. Since the cosmic radiation is the same over time, the concentration of 14 C as Carbon dioxide will be the same in the air over time. There will be a fixed ratio between 12 C and 14 C. The plants take up CO 2 with the ratio. But when the plant die the ratio will be changed when the amount of 14 C decreases due to radioactive decay. By looking at the ratio and knowing the half-life is it possible to determine the age of an object. It accurate to about year old material.

60 Co Penetrating power to treat cancerous cells Gamma (  ) radiation emitter Been used more than 50 years for different cancer forms. Also to stop the immune system to attack transplanted organs.

131 I Half-life of 8 days Beta (  ) and Gamma (  ) emitter Thyroid cancer Diagnose if thyroid gland functions normally

125 I Half-life of 60 days Prostate cancer and brain tumour

Topic 2.2 The mass spectrometer How to measure atomic masses.

1.The sample is Vaporised in vacuum. 2. The sample is Ionised in an electron beam. A + e -  A + + 2e The ions are Accelerated in an electric field into a long tube.

4. The ions are Deflected (= change of flight way) in a magnetic field. 5. Depending of the ions mass and the magnetic field some ion will deflect into a Detector. The number of ion that hit the detector is proportional to the signal from the detector. By changing the power of the magnetic field different ions can be detected.

Concepts to be found and calculated from an mass spectrum How much of the different isotopes of an element is there; i.e. the natural abundance The relative atomic mass of an element You can also calculate the relative atomic mass if you know the natural abundance

A mass spectrometer can be used to determine the natural abundance of isotopes The mass spectrum of Magnesium: The most abundant isotopes are obviously Mg-24, Mg-25 and Mg-26. How many protons and neutrons do these isotopes have? Which mass do they have?

The most abundant isotope is set to be 100. The other ones should be calculated in proportion to this Calculating the total natural abundance, add the numbers on top of the peaks : = Divide each isotopes abundances with the total abundance: 24 Mg =100/127.2 = 78.6% 25 Mg = 12.8/127.2 = 10.0% 26 Mg = 14.4/127.2 = 11.3% These numbers are the natural abundance of the isotopes in Mg

Calculating the relative atomic mass of an element Use the natural abundance of the isotopes in Mg. Multiply the percentage of each isotope with it’s mass: 24* * *0.113 = 24.3 (g/mol)

Exersize: Go to Click on S, sulphur Scroll down to Nuclear properties- click on isotopes How many naturally occurring isotopes does sulphur have? Use the two most abundant to calculate the atomic mass of sulphur.

To calculate the natural abundance from the relative atomic mass You have to know which isotopes are found naturally -Boron, B, has two naturally occurring isotopes, 11 B and 10 B In the periodic table you can see that the relative atomic mass is 10,81 Starting with 100 atoms: Let X atoms be 10 B, then 100-x are 11 B Multiply with the mass and add: 10x+11(100-x) → 10x x→ 1100-x Average mass= total mass/no of atoms: 10,81=(1100-x)/100 → x=19, which means that the abundance of 10 B is 19 % and the abundance of 11 B is 81%

Topic 2.3 Electron arrangement Electromagnetic spectrum Electromagnetic radiation has been very important in the studies of the atom There are different types of electromagnetic radiation: Gamma rays, X-rays, UV, Visible light, IR, Microwaves, Radio waves. They differ in Wavelength. From m to 10 4 m The relation between Wavelength, L(m)and Frequency, n (s -1,Hertz) Ln = cc =speed of light, 3*10 8 m/s Shorter wavelength => Higher frequency =>more energy.

Continuous spectrum

Line spectrum.

+ + Atomíc nucleus K L M Electron shells

Absorptions spectra: energy needed to move an electron to a higher shell. Emission spectra: energy released when the electron falls back to the lower shell. The color of the light show the energy difference between the two shells The longest jump between two neighboring shells is K  L Line converge in the higher frequency end due to many similar energy transitions.

The hydrogen spectrum K L M N The electrons jump ”out” one or more shells- gains energy (excitation) The electrons”falls back” one or more shells- emits energy (relaxation)

The hydrogen spectrum Convergence of lines= ionisation UV radiation Visible radiation IR radiation

Line spectrum of hydrogen (Balmer series)

+ - Electron Hydrogen atom, H 1 p+ K-shell + - Helium atom, He 2 p e Litiumatom, L i 3 p e - K-shell L-shell Berylliumatom, Be 4 p e

+ - B 5 p e - K-shell L-shell Carbon, C 6 p e Nitrogen, N 7 p e Oxygen, O 8 p e Fluor, F 9 p e Neon, Ne 10 p e

ElementAtomic numberKLMN H11 He22 Li321 Be422 B523 C624 N725 O826 Electron arrangement for elements 1-8

Electron arrangement for elements 9-20 ElementAtomic numberKLMN F927 Ne1028 Na11281 Mg12282 Al13283 Si14284 P15285 S16286 Cl17287 Ar18288 K Ca202882

Electron configuration of ions Positive ions (cations) are formed by loss of an electron Electron configuration K K →K + (loss of an electron) K Mg Mg →Mg 2+ (loss of two electrons) Mg Negative ions (anions) are formed by capture of an electron Electron configuration F 2 7 F →F - (capture of an electron) F S S →S 2- (capture of two electrons) S The electrons in the outer shell are called valence electrons