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Unformatted text preview: The Particle in a One Dimensional Box We need to solve the Schrodinger equation for this
problem Flw=Ew where the potential is zero (V=O) inside the box of length
L, but infinite oUtside Of the box. In other words, the
particle is confined to the box. The Hamiltonian (energy)
operator is the same as for the free particle (just kinetic A h?” 62 energy, H = ————2—) and the Schrédinger equation is
2m 6x
7’22 @211}
2m 6x2 w or 6%;
6x2 + which has the general solution A6:ikx + Bea—“0‘ w .80 But now there are boundary conditions. 1st condition: at x=0, w(x=0)=0 w(0)=O=A+B B=—A
So, the general wavefunction becomes
\lJ(X) = A(eimx’t — aim“): ziAsinNTm—Ex/h) = Csin(J2EE’x/
2"“ condition: at x=L, w(x=L)=0, so
ML) = Csin(J2111—EL/h) = 0 the sin function is only zero for whole multiples of TE, so
the 2nd condition is only satisfied when .l2mEnL/h : inn for 11 21.2.3... We have the quantum number n. the system has been quantized by the imposition of a boundary condition.
Solving for En gives 222
Huh
“:2 22130112 E Plugging En back into the general expression for u} gives
Wu 2 iC sin(mtx/L) We can normalize to find C, for either the + or  case fwlwndx = C*Cfsin2(nnx/L)dx =1 Let u=n7cx/L then du=(mr/L)dx, so :C*Cf‘sm2u _I:_.dx=C*C~£a1~(u—sin(u)cos(u) M
1m nrr2 0
1m 2 2 rearranging, C*C=C[2=% so ng therefore for the particleina—box 1pm 2 Esinmnx/L) and nznzhz " Er1 : 2 — Eon2 11 21,2,3...
2mL 0 x A L , . ,0 x  L (a)  _ . .. as) Figu‘re 21 .2 Wave flunc'tions and probability
derisities for_the particle in the box. Ene rgy/(l12/8 mm Suﬁ E 1%..“ E
a , é Fig. 14.2. The allowed energy levels
and the corresponding (sine wave)
wavefunctions for a particle in a box.
Note that the energy levels increase as
n2, and so their separation increases as
the quantum number increases. Each
wavefunction is a standing wave, and
successive functions possess one more
half wave. Probability of Finding the Particle in a Part of the Box _ Problem: Calculate the probability of finding the particle
between 0.2L and 0.3L 3L 2. 2X]. _2'3L.2
PZLJL — lzjh/Esmﬁmx/ LRE sm(mtx/ L)dx — Elstm (mtx/ L)dx
= 2 3 —WI 2 .1 + wl—[sin(.4mt) — sin(.6mt) L 2 4DTC/L .21, znﬂ 1 . 1 2 .194
3 .100
4 .024 w When are guantized energies important? The particlein—a—box energies are 2 2 2
n R h =E0n2 n=1,2,3... En : 2
2mL 5.49xlo—68 st2
: mL2
the mass of the particle and the size of the box. where E0 . So the energy depends on The difference between adjacent energy levels is En+1 —En = EO[(n+1)2 —n2] 2 (1‘12 +2n+1—n2)E0 =(211+1)E(J Consider two cases Case I: a particle the size of a baseball (say mass=0.1
kg) in a box 60’long at a velocity of 90 mph. L=60’(1m/3.281’)=18.3 m
v=90 mph (1609 m/mile)(1 hr/36003)=40 m/s
EKE=O.5(O.1 kg)v2=80 J Eo=5.49x10'68 J232/(o.1 kg (18.3 m)2)=1.63x10'69 J EKE=n2Eo, so 11 = EKF = wigs? :2.2x1035 .
E0 1.63x10 J The spacing between levels is EM] —En = (2n +1)EO = (2 2.2x1035 +1)1.63X10'69 J = 7.2X10'34J to observe quantized levels you would have to resolve an
energy of 7x10'34 J out of 80 J. The levels can be
considered continuous for most purposes. Case ll: an electron (mass=9.1095x‘0'31 kg) in a box the
size of a molecule, say 5x10'10 m or 5 angstroms. E0=5.49x10'68 J2s2/(9.‘l 1x10"31 kg (5x10'10 m)2)
=2.41x10‘19 J= 1.50 eV The spacing between levels is En+1 —EH 2 (2n+1)E0, so for n=1 we have the spacing between the ground and first
excited level as 3E0 which equals E:Z —E1 = 3E0 = 3(1 .50 W) = 4.50eV
This spacing is very large compared to kT (~0.03 eV), so
the fact that the levels are quantized is very important.
The n=1 to n=2 transition energy is well into the ultraviolet. In fact most closed shell molecules have
transitions to their first excited states in the ultraviolet. ...
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 Fall '04
 DR.COE
 Quantum Chemistry

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