how would I find the total energy for n=0,1,2?

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2. If the bond length of HF is 0.91 Å, and frequency is 1.241*1014 s-1 how would I find the total energy for n=0,1,2? 

n = 4
n = 3
-n =D1
n = 0
X = 0
FIGURE 11.4 Plots of the first five wave-
unctions of the harmonic oscillator. They
re superimposed against the potential
nergy for the system. The positions where
ne wavefunctions go outside the poten-
energy are called the classical turning
oints. Classically, a harmonic oscillator will
ever go beyond its turning point, because
does not have enough energy. Quantum
nechanically, there is a nonzero prob-
pility that a particle acting as a harmonic
scillator will exist beyond this point.
al
Transcribed Image Text:n = 4 n = 3 -n =D1 n = 0 X = 0 FIGURE 11.4 Plots of the first five wave- unctions of the harmonic oscillator. They re superimposed against the potential nergy for the system. The positions where ne wavefunctions go outside the poten- energy are called the classical turning oints. Classically, a harmonic oscillator will ever go beyond its turning point, because does not have enough energy. Quantum nechanically, there is a nonzero prob- pility that a particle acting as a harmonic scillator will exist beyond this point. al
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