fail_bohr - ‘\ 41012nm 364.6 nm . '- '4...

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Unformatted text preview: ‘\ 41012nm 364.6 nm . '- '4 "'"-“ -x-_ ‘1'. ;- .u v _ _ l." . . I 4 1 fl --.-1 . . _ . . h- I _l h . - . - . ..’ _ ' '- _ : _ _ ' , . _ E . I . - I I 4-. II I " I a." “I i ' . I .‘ ._ ' 1' _-' _' ' - H." '11 .3 i .I ‘ - ' r . . -.. . j a In ' " - H ‘ - . J 1 .* ‘ t 4 ' .. ' I“ i I I ' In" 0 . * ed ‘ B ue IJVoIet Ultravie 1 I Series limit f-'—-- 1| 3 nm 410.2nm 364.6nm m\ l 4.0 __.iolel: Ultraio 1 ‘ 4 -*--)x- 656.3nm '48 ,3 “tals_‘ll_. Red Blue . f__ Series limit E2 7 Hydrogen Spectrum in Balmer, 1885 . 2 9L=3.6456-10"'7m[ n ] n2—4 Ritz when k=2 this formula reduces to Balmer formula the Rydberg constant (RH) was the most accurately determined constant in physics at that time today RH=1.09677585(8)X107 m'1 Bohr Theory, 1913 Classically, - electrons should spiral into nucleus 0 charge in a circular orbit is accelerating and should radiate Bohr says, 0 electrons can move in certain orbits without radiating, stationary states 0 atoms radiate only when changing from one orbit to another AB = hu = E A o angular momentum is quantized mvr=n_h=nh To stay in orbit there must be a centripetal force of mv2/r This is provided by the electrostatic attraction eZ/(4areor2). So, U 80 is the permittivity of vacuum, 20:8.8545 Fm'1 Cancelling r's 92 4nsor which is the condition for a stable orbit. Bohr assumes that angular momentum (and consequently, the energy and radius) is quantized. mvrzn—h=nh n=‘l,2,3,... 27: Solving this condition for velocity gives mv2 = Plugging the velocity relation into the. condition for a stable orbit gives “iii—i5 92 mr 4n80r which solved for r is 4ns h2n2 r: 0 :8 n2 n=1,2,3... me2 0 where n is a quantum number and ac is the Bohr radius, the atomic unit of distance. a0 =5.29177x10-11m=0.529 A=52.9 pm U To get the energy of each orbit, add the kinetic and potential energy _. _ 1 2 __ e2 E “ Ekinetic + Epotential _ Emv 7'— the negative sign arises because the electrostatic force is attractive between the electron and proton. Substituting for mv2 from the condition for a stable orbit gives E _ 1 e2 e2 1 e2 247:50r 47r80r 247219 r Substituting our quantum relation for r gives E=_ e2 =—2.18x10‘18Jou|es= H 2 2 2 8728080“ n 2n where H is the atomic unit of energy, the Hartree. .. H=27.2‘l eV=627.5 kcaI/moi=4.359814x10'18 J. the Rydberg constant is just H/(2hc). Bohr calculated from theory the best experimentally determined constant in physics. People were impressed. lfthe electron is in the lowest energy orbit, n=1, it is said to be in the ground state. If the electron is in- any orbit greater than n=‘|, it is said to be in an excited state. The ionization energy or ionization potential (IP) is the energy required to remove the electron from the atom. IP:E(n=oo)=E(n=1)=— H + H =fl=13.61ev 2002 2 (12) 2 Absorption or emission lines of hydrogen atom are given by e hc ~ H 1 1 AE=h=—=hc 2— — 0 it D 2 n2 n2 lower higher where mower is less than nmgher because AE must be positive. For emission nhigher is the initial state and mower is the final state. For absorption mower is the initial state and nhigher is the final state. Bohr Theory = Standing Wave Condition Bohr's quantization of angular momentum nh . mvr—fi Can be combined with DeBroglie's relation AzflilL p mv Solve DeBroglie's relation for mv (sh/it) and substitute into Bohr‘s relation to get Brim A 27‘: which rearranges to nk 2 2m Therefore Bohr's stable orbits are the ones whose DeBroglie wavelengths constructively interfere about the circumference of the orbit. The Bohr orbits are DeBroglie standing waves. Another perspective on the Deroglie relation: For photons, Planck gave us E=hv=hclk and Einstein gave us E=mc2. Equating these gives hc/?t=mc2 which rearranges to 9t=h/(mc). For a particle, ' DeBroglie assumed that the particle‘s velocity could be substituted for 0 giving k=h/(mv). H z groundsi. ’M’.‘ ...
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