Unformatted text preview: ‘\ 41012nm 364.6 nm . '- '4 "'"-“
;- .u v _ _
l." . .
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_.
. f__ Series limit
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
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
80 is the permittivity of vacuum, 20:8.8545 Fm'1
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
which solved for r is
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
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 .
Can be combined with DeBroglie's relation
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’.‘ ...
View Full Document